(a) Start with the equation:

\(5 \cos \theta - \sin \theta \tan \theta + 1 = 0\)

Multiply through by \(\cos \theta\) to eliminate \(\tan \theta\):

\(5 \cos^2 \theta - \sin \theta \frac{\sin \theta}{\cos \theta} \cos \theta + \cos \theta = 0\)

Simplify:

\(5 \cos^2 \theta - \sin^2 \theta + \cos \theta = 0\)

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):

\(5 \cos^2 \theta - (1 - \cos^2 \theta) + \cos \theta = 0\)

Simplify further:

\(6 \cos^2 \theta + \cos \theta - 1 = 0\)

Thus, \(a = 6\), \(b = 1\), \(c = -1\).

(b) Solve the quadratic equation:

\((3 \cos \theta - 1)(2 \cos \theta + 1) = 0\)

Set each factor to zero:

\(3 \cos \theta - 1 = 0\) gives \(\cos \theta = \frac{1}{3}\)

\(2 \cos \theta + 1 = 0\) gives \(\cos \theta = -\frac{1}{2}\)

Find \(\theta\) for \(\cos \theta = \frac{1}{3}\):

\(\theta \approx 1.23, 5.05\)

Find \(\theta\) for \(\cos \theta = -\frac{1}{2}\):

\(\theta = \frac{2\pi}{3}, \frac{4\pi}{3}\)

Convert to approximate values:

\(\theta \approx 2.09, 4.19\)

Thus, \(\theta = \{1.23, \frac{2\pi}{3}, 2.09, 4.19, 5.05\}\).

(a) Start with the equation:

\(\frac{1}{\sin \theta + \cos \theta} + \frac{1}{\sin \theta - \cos \theta} = 1\)

Find a common denominator:

\(\frac{\sin \theta - \cos \theta + \sin \theta + \cos \theta}{(\sin \theta + \cos \theta)(\sin \theta - \cos \theta)} = 1\)

Simplify the numerator:

\(\frac{2\sin \theta}{\sin^2 \theta - \cos^2 \theta} = 1\)

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\(2\sin \theta = \sin^2 \theta - (1 - \sin^2 \theta)\)

Simplify to get:

\(2\sin^2 \theta - 2\sin \theta - 1 = 0\)

Thus, the equation is in the form \(a \sin^2 \theta + b \sin \theta + c = 0\) with \(a = 2, b = -2, c = -1\).

(b) Solve the quadratic equation \(2\sin^2 \theta - 2\sin \theta - 1 = 0\) using the quadratic formula:

\(\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substitute \(a = 2, b = -2, c = -1\):

\(\sin \theta = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 2 \times (-1)}}{4}\)

\(\sin \theta = \frac{2 \pm \sqrt{4 + 8}}{4}\)

\(\sin \theta = \frac{2 \pm \sqrt{12}}{4}\)

\(\sin \theta = \frac{2 \pm 2\sqrt{3}}{4}\)

\(\sin \theta = \frac{1 \pm \sqrt{3}}{2}\)

For \(\sin \theta = \frac{1 + \sqrt{3}}{2}\), \(\theta \approx 201.5^\circ\).

For \(\sin \theta = \frac{1 - \sqrt{3}}{2}\), \(\theta \approx 338.5^\circ\).

(a) Let \(u = \sqrt{y}\). Then the equation becomes \(6u + \frac{2}{u} - 7 = 0\).

Multiply through by \(u\) to clear the fraction: \(6u^2 + 2 - 7u = 0\).

Rearrange to form a quadratic: \(6u^2 - 7u + 2 = 0\).

Factor the quadratic: \((2u - 1)(3u - 2) = 0\).

Thus, \(u = \frac{1}{2}\) or \(u = \frac{2}{3}\).

Since \(u = \sqrt{y}\), we have \(\sqrt{y} = \frac{1}{2}\) or \(\sqrt{y} = \frac{2}{3}\).

Therefore, \(y = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) or \(y = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

(b) Use \(\tan x\) in place of \(y\) from part (a).

Thus, \(\tan x = \frac{1}{4}\) or \(\tan x = \frac{4}{9}\).

For \(\tan x = \frac{1}{4}\), \(x = 14.0^\circ\) and \(x = 194.0^\circ\).

For \(\tan x = \frac{4}{9}\), \(x = 24.0^\circ\) and \(x = 204.0^\circ\).

(a) Start with the equation \(4 \cos^4 x + \cos^2 x - 3 = 0\).

Let \(y = \cos^2 x\), then the equation becomes \(4y^2 + y - 3 = 0\).

Factor the quadratic: \((4y - 3)(y + 1) = 0\).

Thus, \(y = \frac{3}{4}\) or \(y = -1\).

Since \(y = \cos^2 x\), \(\cos^2 x = \frac{3}{4}\) (ignore \(y = -1\) as \(\cos^2 x \geq 0\)).

\(\cos x = \pm \frac{\sqrt{3}}{2}\).

Solutions for \(x\) are \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

(b) Use the quadratic formula for \(4 \cos^4 x + \cos^2 x - k = 0\).

Let \(y = \cos^2 x\), then \(4y^2 + y - k = 0\).

Using the quadratic formula: \(y = \frac{-1 \pm \sqrt{1 + 16k}}{8}\).

For \(k > 5\), \(1 + 16k > 81\), so \(\sqrt{1 + 16k} > 9\).

Thus, \(y = \frac{-1 \pm \text{something greater than 9}}{8}\).

This implies \(y < 0\) or \(y > 1\), which are not possible for \(\cos^2 x\).

Therefore, no solutions exist for \(k > 5\).

(a) To prove the identity, start by using a common denominator:

\(\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta} = \frac{\sin^3 \theta (1 + \sin \theta) - \sin^2 \theta (\sin \theta - 1)}{(\sin \theta - 1)(1 + \sin \theta)}\)

Simplify the numerator:

\(\sin^3 \theta + \sin^4 \theta - \sin^3 \theta + \sin^2 \theta = \sin^4 \theta + \sin^2 \theta\)

Factor the numerator:

\(\sin^2 \theta (\sin^2 \theta + 1)\)

Recognize that \(\sin^2 \theta + \cos^2 \theta = 1\), so \(\cos^2 \theta = 1 - \sin^2 \theta\).

Thus, the expression becomes:

\(\frac{\sin^2 \theta (1 + \sin^2 \theta)}{\cos^2 \theta} = \tan^2 \theta (1 + \sin^2 \theta)\)

Therefore, the identity is proven.

(b) To solve the equation:

\(\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta} = \tan^2 \theta (1 - \sin^2 \theta)\)

From part (a), we have:

\(-\tan^2 \theta (1 + \sin^2 \theta) = \tan^2 \theta (1 - \sin^2 \theta)\)

Equating gives:

\(2 \tan^2 \theta = 0\)

Thus, \(\tan^2 \theta = 0\), leading to \(\tan \theta = 0\).

The solutions for \(\tan \theta = 0\) in the interval \(0 < \theta < 2\pi\) are \(\theta = \pi\).

(a) To show the given identity, start by obtaining a common denominator:

\(\frac{(\sin \theta + 2 \cos \theta)(\cos \theta + 2 \sin \theta) - (\sin \theta - 2 \cos \theta)(\cos \theta - 2 \sin \theta)}{(\cos \theta - 2 \sin \theta)(\cos \theta + 2 \sin \theta)}\)

Expanding the numerators:

\(5 \sin \theta \cos \theta + 2 \sin^2 \theta + 2 \cos^2 \theta - (5 \sin \theta \cos \theta - 2 \sin^2 \theta - 2 \cos^2 \theta)\)

\(= 4(\cos^2 \theta + \sin^2 \theta)\)

Using \(\cos^2 \theta + \sin^2 \theta = 1\), the expression simplifies to:

\(\frac{4}{\cos^2 \theta - 4 \sin^2 \theta}\)

\(= \frac{4}{5 \cos^2 \theta - 4}\)

(b) For the equation:

\(\frac{4}{5 \cos^2 \theta - 4} = 5\)

Multiply both sides by \(5 \cos^2 \theta - 4\):

\(4 = 5(5 \cos^2 \theta - 4)\)

\(25 \cos^2 \theta = 24\)

\(\cos^2 \theta = \frac{24}{25}\)

\(\cos \theta = \pm \sqrt{\frac{24}{25}} = \pm 0.9798\)

For \(0^\circ < \theta < 180^\circ\), \(\theta = 11.5^\circ\) or \(168.5^\circ\).

(a) Start with \(\frac{\tan x + \cos x}{\tan x - \cos x} = k\).

Use \(\tan x = \frac{\sin x}{\cos x}\) to rewrite the equation: \(\frac{\frac{\sin x}{\cos x} + \cos x}{\frac{\sin x}{\cos x} - \cos x} = k\).

Clear the fraction: \(\frac{\sin x + \cos^2 x}{\sin x - \cos^2 x} = k\).

Use \(\cos^2 x = 1 - \sin^2 x\) to get \(\frac{\sin x + 1 - \sin^2 x}{\sin x - (1 - \sin^2 x)} = k\).

Simplify: \(\frac{\sin x + 1 - \sin^2 x}{\sin^2 x + \sin x - 1} = k\).

Multiply through by the denominator: \((\sin x + 1 - \sin^2 x) = k(\sin^2 x + \sin x - 1)\).

Expand and simplify: \(\sin x + 1 - \sin^2 x = k\sin^2 x + k\sin x - k\).

Rearrange: \((k + 1)\sin^2 x + (k - 1)\sin x - (k + 1) = 0\).

(b) Solve \(\frac{\tan x + \cos x}{\tan x - \cos x} = 4\).

Using the result from part (a), substitute \(k = 4\): \(5\sin^2 x + 3\sin x - 5 = 0\).

Use the quadratic formula: \(\sin x = \frac{-3 \pm \sqrt{9 + 100}}{10}\).

Calculate: \(\sin x = \frac{-3 \pm 11}{10}\).

Solutions: \(\sin x = 0.8\) or \(\sin x = -1.4\) (discard \(\sin x = -1.4\) as it is not possible).

Find \(x\) for \(\sin x = 0.8\): \(x = 48.1^\circ, 131.9^\circ\).

Start by multiplying through by \(\cos \theta\) to eliminate the fraction:

\(2 \cos^2 \theta = 7 \cos \theta - 3\)

Rearrange to form a quadratic equation:

\(2 \cos^2 \theta - 7 \cos \theta + 3 = 0\)

Factor the quadratic equation:

\((2 \cos \theta - 1)(\cos \theta - 3) = 0\)

Set each factor to zero and solve for \(\cos \theta\):

\(2 \cos \theta - 1 = 0\) gives \(\cos \theta = \frac{1}{2}\)

\(\cos \theta - 3 = 0\) gives \(\cos \theta = 3\) (not possible since \(|\cos \theta| \leq 1\))

For \(\cos \theta = \frac{1}{2}\), \(\theta = \pm 60^\circ\) within the given range.

Thus, \(\theta = -60^\circ\) and \(\theta = 60^\circ\).

(a) Start with the identity \(1 - \tan^2 \theta\).

We know \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\), so \(1 - \tan^2 \theta = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}\).

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), we have:

\(\frac{1 - 2\sin^2 \theta}{1 - \sin^2 \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta} = 1 - \tan^2 \theta\).

(b) Given \(\frac{1 - 2 \sin^2 \theta}{1 - \sin^2 \theta} = 2 \tan^4 \theta\), substitute \(1 - \tan^2 \theta = 2 \tan^4 \theta\).

This simplifies to \(1 - \tan^2 \theta = 2 \tan^4 \theta \Rightarrow 2 \tan^4 \theta + \tan^2 \theta - 1 = 0\).

Let \(x = \tan^2 \theta\), then \(2x^2 + x - 1 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = 1, c = -1\):

\(x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\).

Thus, \(x = 0.5\) or \(x = -1\). Since \(x = \tan^2 \theta\), \(x = -1\) is not valid.

So, \(\tan^2 \theta = 0.5\) leading to \(\tan \theta = \pm \sqrt{0.5}\).

\(\theta = 35.3^\circ\) and \(\theta = 144.7^\circ\) (A.W.R.T).

Start with the equation:

\(3 \tan^2 \theta + 1 = \frac{2}{\tan^2 \theta}\)

Multiply through by \(\tan^2 \theta\) to eliminate the fraction:

\(3 \tan^4 \theta + \tan^2 \theta - 2 = 0\)

Let \(x = \tan^2 \theta\), then the equation becomes:

\(3x^2 + x - 2 = 0\)

Factor the quadratic equation:

\((3x - 2)(x + 1) = 0\)

Thus, \(x = \frac{2}{3}\) or \(x = -1\).

Since \(x = \tan^2 \theta\), \(x\) must be non-negative, so \(x = -1\) is not valid.

Therefore, \(\tan^2 \theta = \frac{2}{3}\).

Take the square root to find \(\tan \theta\):

\(\tan \theta = \pm \sqrt{\frac{2}{3}}\)

Calculate \(\theta\) using the inverse tangent function:

\(\theta = \tan^{-1}(\sqrt{\frac{2}{3}}) \approx 39.2^\circ\)

Since \(\tan \theta\) is positive in the first and third quadrants, the solutions are:

\(\theta = 39.2^\circ\) and \(\theta = 180^\circ - 39.2^\circ = 140.8^\circ\)

(i) Start with the equation \(3 \cos^4 \theta + 4 \sin^2 \theta - 3 = 0\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\).

Substitute \(\sin^2 \theta = 1 - x\) where \(x = \cos^2 \theta\).

The equation becomes \(3x^2 + 4(1-x) - 3 = 0\).

Simplify to get \(3x^2 - 4x + 1 = 0\).

(ii) Solve the quadratic equation \(3x^2 - 4x + 1 = 0\).

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -4\), \(c = 1\).

Calculate the discriminant: \((-4)^2 - 4 \times 3 \times 1 = 4\).

\(x = \frac{4 \pm 2}{6}\), giving \(x = 1\) or \(x = \frac{1}{3}\).

For \(x = 1\), \(\cos^2 \theta = 1\), so \(\cos \theta = \pm 1\).

\(\theta = 0^\circ, 180^\circ\).

For \(x = \frac{1}{3}\), \(\cos^2 \theta = \frac{1}{3}\), so \(\cos \theta = \pm \frac{1}{\sqrt{3}}\).

\(\theta = 54.7^\circ, 125.3^\circ\).

(a) Expand \((2x - 1)(4x^2 + 2x - 1)\):

\((2x - 1)(4x^2 + 2x - 1) = 8x^3 + 4x^2 - 2x - 4x^2 - 2x + 1 = 8x^3 - 4x + 1\).

Thus, the identity is verified.

(b) Start with the left-hand side:

\(\frac{\tan^2 \theta + 1}{\tan^2 \theta - 1} = \frac{\frac{\sin^2 \theta}{\cos^2 \theta} + 1}{\frac{\sin^2 \theta}{\cos^2 \theta} - 1} = \frac{\frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta}}{\frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta}} = \frac{1}{1 - 2\cos^2 \theta}\).

Thus, the identity is proven.

(c) Using the results of (a) and (b), solve:

\(\frac{1}{1 - 2\cos^2 \theta} = 4\cos \theta\).

Cross-multiply to get:

\(1 = 4\cos \theta (1 - 2\cos^2 \theta)\).

Simplify to:

\(4\cos \theta - 8\cos^3 \theta = 1\).

Let \(x = \cos \theta\), then:

\(8x^3 - 4x + 1 = 0\).

Using the identity from (a), solve for \(x\):

\((2x - 1)(4x^2 + 2x - 1) = 0\).

\(2x - 1 = 0\) gives \(x = \frac{1}{2}\), \(\cos \theta = \frac{1}{2}\), \(\theta = 60^\circ\).

\(4x^2 + 2x - 1 = 0\) gives \(x = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm 4}{8}\).

\(x = \frac{1}{4}\), \(\cos \theta = \frac{1}{4}\), \(\theta = 75.5^\circ\) (not in range).

\(x = -\frac{3}{4}\), \(\cos \theta = -\frac{3}{4}\), \(\theta = 144^\circ\).

Thus, \(\theta = 60^\circ, 72^\circ, 144^\circ\).

(i) Start with the equation:

\(4 \tan x + 3 \cos x + \frac{1}{\cos x} = 0\)

Multiply through by \(\cos x\) to eliminate the fraction:

\(4 \sin x + 3 \cos^2 x + 1 = 0\)

Use the identity \(\cos^2 x = 1 - \sin^2 x\):

\(4 \sin x + 3(1 - \sin^2 x) + 1 = 0\)

Simplify to form a quadratic in \(\sin x\):

\(3 \sin^2 x - 4 \sin x - 4 = 0\)

Solving this quadratic gives \(\sin x = -\frac{2}{3}\).

(ii) Substitute \(\sin(2x - 20^\circ) = -\frac{2}{3}\) into the equation:

\(2x - 20^\circ = 221.8^\circ, 318.2^\circ\)

Solve for \(x\):

\(x = 120.9^\circ, 169.1^\circ\)

Start with the equation:

\(3 \sin^2 2\theta + 8 \cos 2\theta = 0\)

Use the identity \(\sin^2 2\theta = 1 - \cos^2 2\theta\) to rewrite the equation:

\(3(1 - \cos^2 2\theta) + 8 \cos 2\theta = 0\)

Simplify to:

\(3 - 3\cos^2 2\theta + 8 \cos 2\theta = 0\)

Rearrange to form a quadratic equation in \(\cos 2\theta\):

\(3\cos^2 2\theta - 8 \cos 2\theta - 3 = 0\)

Let \(x = \cos 2\theta\), then solve the quadratic:

\(3x^2 - 8x - 3 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -8\), \(c = -3\):

\(x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times (-3)}}{2 \times 3}\)

\(x = \frac{8 \pm \sqrt{64 + 36}}{6}\)

\(x = \frac{8 \pm \sqrt{100}}{6}\)

\(x = \frac{8 \pm 10}{6}\)

\(x = 3.0 \text{ or } x = -\frac{1}{3}\)

Since \(\cos 2\theta\) must be between -1 and 1, we take \(\cos 2\theta = -\frac{1}{3}\).

Find \(2\theta\):

\(2\theta = \cos^{-1}\left(-\frac{1}{3}\right)\)

\(2\theta \approx 109.47^\circ \text{ or } 250.53^\circ\)

Thus, \(\theta \approx 54.7^\circ \text{ or } 125.3^\circ\).

(i) Start with the expression:

\(\frac{\tan \theta + 1}{1 + \cos \theta} + \frac{\tan \theta - 1}{1 - \cos \theta}\)

Combine the fractions:

\(\frac{(\tan \theta + 1)(1 - \cos \theta) + (\tan \theta - 1)(1 + \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)}\)

Simplify the numerator:

\(\tan \theta - \tan \theta \cos \theta + 1 - \cos \theta + \tan \theta + \tan \theta \cos \theta - 1 - \cos \theta\)

\(= 2\tan \theta - 2\cos \theta\)

The denominator is:

\(1 - \cos^2 \theta = \sin^2 \theta\)

Thus, the expression simplifies to:

\(\frac{2(\tan \theta - \cos \theta)}{\sin^2 \theta}\)

(ii) Set the expression to zero:

\(\frac{2(\tan \theta - \cos \theta)}{\sin^2 \theta} = 0\)

This implies:

\(\tan \theta - \cos \theta = 0\)

\(\frac{\sin \theta}{\cos \theta} = \cos \theta\)

\(\sin \theta = \cos^2 \theta\)

\(\sin \theta = 1 - \sin^2 \theta\)

\(\sin^2 \theta + \sin \theta - 1 = 0\)

Using the quadratic formula, solve for \(\sin \theta\):

\(\sin \theta = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\)

\(\sin \theta = 0.618\)

\(\theta = 38.2^\circ\)

(i) Start with the equation:

\(\frac{\cos \theta - 4}{\sin \theta} - \frac{4 \sin \theta}{5 \cos \theta - 2} = 0\)

Multiply through by \(\sin \theta (5 \cos \theta - 2)\) to clear the denominators:

\((\cos \theta - 4)(5 \cos \theta - 2) - 4 \sin^2 \theta = 0\)

Expand and simplify:

\(5 \cos^2 \theta - 22 \cos \theta + 8 - 4(1 - \cos^2 \theta) = 0\)

\(5 \cos^2 \theta - 22 \cos \theta + 8 - 4 + 4 \cos^2 \theta = 0\)

Combine like terms:

\(9 \cos^2 \theta - 22 \cos \theta + 4 = 0\)

Thus, the equation is shown as required.

(ii) Solve the quadratic equation:

\(9 \cos^2 \theta - 22 \cos \theta + 4 = 0\)

Using the quadratic formula \(\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9\), \(b = -22\), \(c = 4\):

\(\cos \theta = \frac{22 \pm \sqrt{(-22)^2 - 4 \times 9 \times 4}}{18}\)

\(\cos \theta = \frac{22 \pm \sqrt{484 - 144}}{18}\)

\(\cos \theta = \frac{22 \pm \sqrt{340}}{18}\)

\(\cos \theta = \frac{22 \pm 18.44}{18}\)

\(\cos \theta = 0.1978 \text{ or } 2.247\)

Since \(\cos \theta\) must be between -1 and 1, only \(\cos \theta = 0.1978\) is valid.

Find \(\theta\):

\(\theta = \cos^{-1}(0.1978) \approx 78.6^\circ\)

For the second solution in the range \(0^\circ \leq \theta \leq 360^\circ\), use \(360^\circ - 78.6^\circ = 281.4^\circ\).

Thus, \(\theta = 78.6^\circ, 281.4^\circ\).

Start by multiplying both sides by the denominator \(3 + 2 \tan x\) to eliminate the fraction:

\(5 + 2 \tan x = (1 + \tan x)(3 + 2 \tan x)\)

Expand the right side:

\(5 + 2 \tan x = 3 + 3 \tan x + 2 \tan x + 2 \tan^2 x\)

Combine like terms:

\(5 + 2 \tan x = 3 + 5 \tan x + 2 \tan^2 x\)

Rearrange to form a quadratic equation:

\(2 \tan^2 x + 3 \tan x - 2 = 0\)

Use the quadratic formula \(\tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = 3\), \(c = -2\):

\(\tan x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}\)

\(\tan x = \frac{-3 \pm \sqrt{9 + 16}}{4}\)

\(\tan x = \frac{-3 \pm 5}{4}\)

\(\tan x = \frac{2}{4} = 0.5\) or \(\tan x = \frac{-8}{4} = -2\)

For \(\tan x = 0.5\), \(x = \tan^{-1}(0.5) \approx 0.464\) (accept \(0.148\pi\))

For \(\tan x = -2\), \(x = \tan^{-1}(-2) \approx 2.03\) (accept \(0.648\pi\))

(i) Start with the equation:

\(\frac{\cos \theta + 4}{\sin \theta + 1} + 5 \sin \theta - 5 = 0\)

Multiply throughout by \(\sin \theta + 1\):

\(\cos \theta + 4 + 5 \sin \theta (\sin \theta + 1) - 5(\sin \theta + 1) = 0\)

\(\cos \theta + 4 + 5 \sin^2 \theta + 5 \sin \theta - 5 \sin \theta - 5 = 0\)

\(\cos \theta + 4 + 5 \sin^2 \theta - 5 = 0\)

\(\cos \theta - 1 + 5 \sin^2 \theta = 0\)

Use \(\sin^2 \theta = 1 - \cos^2 \theta\):

\(\cos \theta - 1 + 5(1 - \cos^2 \theta) = 0\)

\(\cos \theta - 1 + 5 - 5 \cos^2 \theta = 0\)

\(5 \cos^2 \theta - \cos \theta - 4 = 0\)

(ii) Solve \(5 \cos^2 \theta - \cos \theta - 4 = 0\).

Factorize or use the quadratic formula:

\(\cos \theta = 1 \quad \text{and} \quad \cos \theta = -0.8\)

For \(\cos \theta = 1\), \(\theta = 0^\circ, 360^\circ\).

For \(\cos \theta = -0.8\), \(\theta = 143.1^\circ, 216.9^\circ\).

Thus, \(\theta = [0^\circ, 360^\circ], [143.1^\circ], [216.9^\circ]\).

(i) Start with the equation \(\cos 2x(\tan^2 2x + 3) + 3 = 0\).

Use the identity \(\tan^2 2x = \frac{\sin^2 2x}{\cos^2 2x}\).

Replace \(\sin^2 2x\) with \(1 - \cos^2 2x\):

\(\tan^2 2x = \frac{1 - \cos^2 2x}{\cos^2 2x} = \sec^2 2x - 1\).

Substitute back into the equation:

\(\cos 2x((\sec^2 2x - 1) + 3) + 3 = 0\).

Simplify to get:

\(\cos 2x(\sec^2 2x + 2) + 3 = 0\).

Multiply through by \(\cos^2 2x\) and rearrange:

\(2 \cos^2 2x + 3 \cos 2x + 1 = 0\).

(ii) Solve \(2 \cos^2 2x + 3 \cos 2x + 1 = 0\).

Let \(y = \cos 2x\), then the equation becomes:

\(2y^2 + 3y + 1 = 0\).

Factorize to find \(y = -\frac{1}{2}\) or \(y = -1\).

For \(\cos 2x = -\frac{1}{2}\), \(2x = 120^\circ\) or \(240^\circ\), so \(x = 60^\circ\) or \(120^\circ\).

For \(\cos 2x = -1\), \(2x = 180^\circ\), so \(x = 90^\circ\).

Thus, the solutions are \(x = 60^\circ, 90^\circ, 120^\circ\).

(i) Expand the left-hand side: \((\sin \theta + 2 \cos \theta)(1 + \sin \theta - \cos \theta) = \sin \theta + \sin^2 \theta - \sin \theta \cos \theta + 2 \cos \theta + 2 \cos \theta \sin \theta - 2 \cos^2 \theta\).

Rearrange terms: \(\sin \theta + \sin^2 \theta - \sin \theta \cos \theta + 2 \cos \theta + 2 \cos \theta \sin \theta - 2 \cos^2 \theta = \sin \theta + 2 \cos \theta + \sin^2 \theta + \sin \theta \cos \theta - 2 \cos^2 \theta\).

Use \(\sin^2 \theta = 1 - \cos^2 \theta\): \(\sin \theta + 2 \cos \theta + (1 - \cos^2 \theta) + \sin \theta \cos \theta - 2 \cos^2 \theta = \sin \theta + 2 \cos \theta + 1 - \cos^2 \theta + \sin \theta \cos \theta - 2 \cos^2 \theta\).

Simplify: \(1 + \sin \theta + 2 \cos \theta - 3 \cos^2 \theta + \sin \theta \cos \theta = 0\).

Rearrange: \(3 \cos^2 \theta - 2 \cos \theta - 1 = 0\).

(ii) Solve \(3 \cos^2 \theta - 2 \cos \theta - 1 = 0\).

Let \(x = \cos \theta\), then \(3x^2 - 2x - 1 = 0\).

Factorize: \((3x + 1)(x - 1) = 0\).

Solutions: \(x = 1\) or \(x = -\frac{1}{3}\).

For \(\cos \theta = 1\), \(\theta = 0^\circ\).

For \(\cos \theta = -\frac{1}{3}\), \(\theta = 109.5^\circ\) or \(\theta = -109.5^\circ\).

(i) Start with the identity \(\cos^2 x = 1 - \sin^2 x\).

Then, \(\cos^4 x = (\cos^2 x)^2 = (1 - \sin^2 x)^2\).

Expanding, \((1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x\).

Thus, \(\cos^4 x = 1 - 2\sin^2 x + \sin^4 x\).

(ii) Substitute \(\cos^4 x = 1 - 2\sin^2 x + \sin^4 x\) into the equation:

\(8\sin^4 x + (1 - 2\sin^2 x + \sin^4 x) = 2(1 - \sin^2 x)\).

Simplify to get \(9\sin^4 x + 1 - 2\sin^2 x = 2 - 2\sin^2 x\).

Thus, \(9\sin^4 x = 1\).

\(\sin^4 x = \frac{1}{9}\).

\(\sin^2 x = \frac{1}{3}\).

\(\sin x = \pm \sqrt{\frac{1}{3}}\).

Solving for \(x\) gives \(x = 35.3^\circ, 144.7^\circ, 215.3^\circ, 324.7^\circ\).

(i) Start with the equation \(3 \sin x \tan x - \cos x + 1 = 0\). Using \(\tan x = \frac{\sin x}{\cos x}\), rewrite the equation as \(3 \sin^2 x - \cos^2 x + \cos x = 0\). Substitute \(\sin^2 x = 1 - \cos^2 x\) to get \(3(1 - \cos^2 x) - \cos^2 x + \cos x = 0\). Simplify to \(3 - 3\cos^2 x - \cos^2 x + \cos x = 0\), which becomes \(4\cos^2 x - \cos x - 3 = 0\). Solve the quadratic equation \(4c^2 - c - 3 = 0\) for \(c = \cos x\). The solutions are \(\cos x = -\frac{3}{4}\) and \(\cos x = 1\). For \(\cos x = -\frac{3}{4}\), \(x = 2.42\) (allow \(0.77\pi\)). For \(\cos x = 1\), \(x = 0\).

(ii) The equation is \(3 \sin 2x \tan 2x - \cos 2x + 1 = 0\). Using \(\tan 2x = \frac{\sin 2x}{\cos 2x}\), rewrite as \(3 \sin^2 2x - \cos^2 2x + \cos 2x = 0\). Substitute \(\sin^2 2x = 1 - \cos^2 2x\) to get \(3(1 - \cos^2 2x) - \cos^2 2x + \cos 2x = 0\). Simplify to \(4\cos^2 2x - \cos 2x - 3 = 0\). Solve the quadratic equation \(4c^2 - c - 3 = 0\) for \(c = \cos 2x\). The solutions are \(2x = 2\pi - 2.42\) or \(360 - 138.6\). Thus, \(x = 1.21\) (\(0.385\pi\)), \(1.93\) (\(0.614\pi\)), \(0\), \(\pi\) (\(3.14\)).

(a) Start with the equation:

\(4 \sin x + \frac{5}{\tan x} + \frac{2}{\sin x} = 0\)

Multiply through by \(\sin x\) to eliminate the fractions:

\(4 \sin^2 x + 5 \cos x + 2 = 0\)

Use the identity \(\sin^2 x = 1 - \cos^2 x\):

\(4(1 - \cos^2 x) + 5 \cos x + 2 = 0\)

Simplify to get:

\(4 - 4 \cos^2 x + 5 \cos x + 2 = 0\)

Rearrange to form a quadratic in \(\cos x\):

\(4 \cos^2 x - 5 \cos x - 6 = 0\)

Thus, \(a = 4, b = -5, c = -6\).

(b) Solve the quadratic equation:

\(4 \cos^2 x - 5 \cos x - 6 = 0\)

Factorize:

\((4 \cos x + 3)(\cos x - 2) = 0\)

Set each factor to zero:

\(4 \cos x + 3 = 0\) or \(\cos x - 2 = 0\)

\(\cos x = -\frac{3}{4}\) or \(\cos x = 2\)

Since \(\cos x = 2\) is not possible, solve \(\cos x = -\frac{3}{4}\):

\(x = \cos^{-1}(-\frac{3}{4})\)

\(x \approx 138.6^\circ\) and \(x \approx 221.4^\circ\)

Start with the equation \(3 \sin^2 \theta = 4 \cos \theta - 1\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to rewrite the equation:

\(3(1 - \cos^2 \theta) = 4 \cos \theta - 1\)

Expand and simplify:

\(3 - 3 \cos^2 \theta = 4 \cos \theta - 1\)

Rearrange to form a quadratic equation in \(\cos \theta\):

\(3 \cos^2 \theta + 4 \cos \theta - 4 = 0\)

Let \(c = \cos \theta\). The equation becomes:

\(3c^2 + 4c - 4 = 0\)

Use the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 3\), \(b = 4\), \(c = -4\):

\(c = \frac{-4 \pm \sqrt{4^2 - 4 \times 3 \times (-4)}}{2 \times 3}\)

\(c = \frac{-4 \pm \sqrt{16 + 48}}{6}\)

\(c = \frac{-4 \pm \sqrt{64}}{6}\)

\(c = \frac{-4 \pm 8}{6}\)

Solutions for \(c\) are \(c = \frac{2}{3}\) or \(c = -2\).

Since \(\cos \theta\) must be between -1 and 1, \(c = -2\) is not valid.

Thus, \(\cos \theta = \frac{2}{3}\).

Find \(\theta\) using \(\cos^{-1}\):

\(\theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ\)

The second solution in the range \(0^\circ \leq \theta \leq 360^\circ\) is:

\(\theta = 360^\circ - 48.2^\circ = 311.8^\circ\)

Therefore, \(\theta = 48.2^\circ\) or \(\theta = 311.8^\circ\).

Start with the equation:

\(\frac{1}{\cos \theta} + 3 \sin \theta \tan \theta + 4 = 0\)

Multiply through by \(\cos \theta\) to eliminate the fraction:

\(1 + 3 \sin^2 \theta + 4 \cos \theta = 0\)

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):

\(1 + 3(1 - \cos^2 \theta) + 4 \cos \theta = 0\)

Simplify:

\(1 + 3 - 3 \cos^2 \theta + 4 \cos \theta = 0\)

\(3 \cos^2 \theta - 4 \cos \theta - 4 = 0\)

Now solve \(3 \cos^2 \theta - 4 \cos \theta - 4 = 0\):

Let \(x = \cos \theta\), then:

\(3x^2 - 4x - 4 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}\)

\(x = \frac{4 \pm \sqrt{16 + 48}}{6}\)

\(x = \frac{4 \pm \sqrt{64}}{6}\)

\(x = \frac{4 \pm 8}{6}\)

\(x = 2 \text{ or } x = -\frac{2}{3}\)

Since \(\cos \theta\) cannot be 2, use \(\cos \theta = -\frac{2}{3}\):

\(\theta = \cos^{-1}(-\frac{2}{3})\)

\(\theta \approx 131.8^\circ \text{ or } 228.2^\circ\)

(i) Start with the equation \(\frac{4 \cos \theta}{\tan \theta} + 15 = 0\).

Replace \(\tan \theta\) with \(\frac{\sin \theta}{\cos \theta}\), giving \(\frac{4 \cos \theta}{\frac{\sin \theta}{\cos \theta}} + 15 = 0\).

This simplifies to \(\frac{4 \cos^2 \theta}{\sin \theta} + 15 = 0\).

Multiply through by \(\sin \theta\) to clear the fraction: \(4 \cos^2 \theta + 15 \sin \theta = 0\).

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to get \(4(1 - \sin^2 \theta) + 15 \sin \theta = 0\).

Expand and rearrange: \(4 - 4 \sin^2 \theta + 15 \sin \theta = 0\).

Rearrange to \(4 \sin^2 \theta - 15 \sin \theta - 4 = 0\), as required.

(ii) Solve \(4 \sin^2 \theta - 15 \sin \theta - 4 = 0\).

Let \(s = \sin \theta\). The equation becomes \(4s^2 - 15s - 4 = 0\).

Use the quadratic formula: \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -15\), \(c = -4\).

Calculate the discriminant: \(b^2 - 4ac = (-15)^2 - 4 \times 4 \times (-4) = 225 + 64 = 289\).

\(s = \frac{15 \pm \sqrt{289}}{8} = \frac{15 \pm 17}{8}\).

This gives \(s = 4\) or \(s = -\frac{1}{4}\).

Since \(\sin \theta\) must be between -1 and 1, \(s = 4\) is not valid.

Thus, \(\sin \theta = -\frac{1}{4}\).

Find \(\theta\) using \(\sin^{-1}(-\frac{1}{4})\), which gives \(\theta = 194.5^\circ\) or \(\theta = 345.5^\circ\) within the range \(0^\circ \leq \theta \leq 360^\circ\).

(i) Start with \(\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta}\).

Divide the numerator and the denominator by \(\cos \theta\):

\(\frac{\frac{\sin \theta}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta} + 1} = \frac{\tan \theta - 1}{\tan \theta + 1}\).

This proves the identity.

(ii) Using the identity from part (i), \(\frac{\tan \theta - 1}{\tan \theta + 1} = \frac{\tan \theta}{6}\).

Let \(t = \tan \theta\), then \(\frac{t - 1}{t + 1} = \frac{t}{6}\).

Cross-multiply to get:

\(6(t - 1) = t(t + 1)\).

Expand and simplify:

\(6t - 6 = t^2 + t\).

Rearrange to form a quadratic equation:

\(t^2 - 5t + 6 = 0\).

Factorize the quadratic:

\((t - 2)(t - 3) = 0\).

So, \(t = 2\) or \(t = 3\).

Thus, \(\tan \theta = 2\) or \(\tan \theta = 3\).

Find \(\theta\) using inverse tangent:

\(\theta = \tan^{-1}(2) \approx 63.4^\circ\) or \(\theta = \tan^{-1}(3) \approx 71.6^\circ\).

(i) Start with the equation \(1 + \sin x \tan x = 5 \cos x\).

Replace \(\tan x\) with \(\frac{\sin x}{\cos x}\):

\(1 + \frac{\sin^2 x}{\cos x} = 5 \cos x\)

Multiply through by \(\cos x\) to eliminate the fraction:

\(\cos x + \sin^2 x = 5 \cos^2 x\)

Use the identity \(\sin^2 x = 1 - \cos^2 x\):

\(\cos x + 1 - \cos^2 x = 5 \cos^2 x\)

Rearrange to form a quadratic equation:

\(6 \cos^2 x - \cos x - 1 = 0\)

(ii) Solve the quadratic equation \(6 \cos^2 x - \cos x - 1 = 0\).

Using the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = -1\), \(c = -1\):

\(\cos x = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 6 \times (-1)}}{2 \times 6}\)

\(\cos x = \frac{1 \pm \sqrt{1 + 24}}{12}\)

\(\cos x = \frac{1 \pm 5}{12}\)

\(\cos x = \frac{6}{12} = \frac{1}{2}\) or \(\cos x = \frac{-4}{12} = -\frac{1}{3}\)

For \(\cos x = \frac{1}{2}\), \(x = 60^\circ\).

For \(\cos x = -\frac{1}{3}\), \(x \approx 109.5^\circ\).

(i) Start with the equation \(4 \sin^2 x + 8 \cos x - 7 = 0\). Use the identity \(\sin^2 x = 1 - \cos^2 x\) to rewrite the equation as:

\(4(1 - \cos^2 x) + 8 \cos x - 7 = 0\)

Simplify to get:

\(4 - 4 \cos^2 x + 8 \cos x - 7 = 0\)

\(-4 \cos^2 x + 8 \cos x - 3 = 0\)

Let \(c = \cos x\), then:

\(4c^2 - 8c + 3 = 0\)

Factor the quadratic equation:

\((2c - 1)(2c - 3) = 0\)

Thus, \(c = \frac{1}{2}\) or \(c = \frac{3}{2}\). Since \(\cos x\) must be between -1 and 1, \(c = \frac{3}{2}\) is not valid.

For \(c = \frac{1}{2}\), \(\cos x = \frac{1}{2}\) gives \(x = 60^\circ\) or \(x = 300^\circ\).

(ii) For \(4 \sin^2 \left(\frac{1}{2} \theta\right) + 8 \cos \left(\frac{1}{2} \theta\right) - 7 = 0\), let \(x = \frac{1}{2} \theta\).

From part (i), \(x = 60^\circ\) or \(x = 300^\circ\).

Thus, \(\frac{1}{2} \theta = 60^\circ\) or \(\frac{1}{2} \theta = 300^\circ\).

Therefore, \(\theta = 120^\circ\) or \(\theta = 600^\circ\). Since \(\theta\) must be between \(0^\circ\) and \(360^\circ\), \(\theta = 120^\circ\) is the only valid solution.

(i) Start with the equation \(2 \cos^2 \theta = \tan^2 \theta\).

We know \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\).

Thus, \(2 \cos^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\).

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have \(\sin^2 \theta = 1 - \cos^2 \theta\).

Substitute to get \(2 \cos^4 \theta = 1 - \cos^2 \theta\).

Rearrange to form a quadratic: \(2c^4 - c^2 - 1 = 0\), where \(c = \cos \theta\).

(ii) Solve the quadratic equation \((2c^2 - 1)(c^2 + 1) = 0\).

This gives \(c^2 = \frac{1}{2}\) or \(c^2 = -1\).

Since \(c^2 = -1\) has no real solutions, we consider \(c^2 = \frac{1}{2}\).

Thus, \(c = \pm \frac{1}{\sqrt{2}}\).

For \(c = \frac{1}{\sqrt{2}}\), \(\theta = \frac{\pi}{4}\).

For \(c = -\frac{1}{\sqrt{2}}\), \(\theta = \frac{3\pi}{4}\).

Therefore, the solutions are \(\theta = \frac{\pi}{4}\) or \(\frac{3\pi}{4}\).

Start with the equation:

\(7 \cos x + 5 = 2 \sin^2 x\)

Use the identity \(\sin^2 x = 1 - \cos^2 x\):

\(7 \cos x + 5 = 2(1 - \cos^2 x)\)

Expand and rearrange:

\(7 \cos x + 5 = 2 - 2 \cos^2 x\)

\(2 \cos^2 x + 7 \cos x + 3 = 0\)

Factor the quadratic equation:

\((2 \cos x + 1)(\cos x + 3) = 0\)

Set each factor to zero:

\(2 \cos x + 1 = 0\) or \(\cos x + 3 = 0\)

\(\cos x = -\frac{1}{2}\) or \(\cos x = -3\)

Since \(\cos x = -3\) is not possible, solve \(\cos x = -\frac{1}{2}\):

\(x = 120^\circ, 240^\circ\)

(i) Start with the equation \(2 \cos x = 3 \tan x\). Replace \(\tan x\) with \(\frac{\sin x}{\cos x}\), giving \(2 \cos x = 3 \frac{\sin x}{\cos x}\).

Multiply through by \(\cos x\) to eliminate the fraction: \(2 \cos^2 x = 3 \sin x\).

Use the identity \(\cos^2 x = 1 - \sin^2 x\) to rewrite the equation: \(2(1 - \sin^2 x) = 3 \sin x\).

Expand and rearrange: \(2 - 2 \sin^2 x = 3 \sin x\).

Rearrange to form a quadratic equation: \(2 \sin^2 x + 3 \sin x - 2 = 0\).

(ii) Start with the equation \(2 \cos 2y = 3 \tan 2y\). Replace \(\tan 2y\) with \(\frac{\sin 2y}{\cos 2y}\), giving \(2 \cos 2y = 3 \frac{\sin 2y}{\cos 2y}\).

Multiply through by \(\cos 2y\) to eliminate the fraction: \(2 \cos^2 2y = 3 \sin 2y\).

Use the identity \(\cos^2 2y = 1 - \sin^2 2y\) to rewrite the equation: \(2(1 - \sin^2 2y) = 3 \sin 2y\).

Expand and rearrange: \(2 - 2 \sin^2 2y = 3 \sin 2y\).

Rearrange to form a quadratic equation: \(2 \sin^2 2y + 3 \sin 2y - 2 = 0\).

Solve the quadratic equation for \(\sin 2y\): \(\sin 2y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}\).

Calculate the discriminant: \(9 + 16 = 25\).

\(\sin 2y = \frac{-3 \pm 5}{4}\).

\(\sin 2y = \frac{2}{4} = \frac{1}{2}\) or \(\sin 2y = \frac{-8}{4} = -2\) (not possible).

For \(\sin 2y = \frac{1}{2}\), \(2y = 30^\circ\) or \(2y = 150^\circ\).

Thus, \(y = 15^\circ\) or \(y = 75^\circ\).

(i) Start with the equation \(2 \cos^2 \theta = 3 \sin \theta\). Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), rewrite the equation as:

\(2(1 - \sin^2 \theta) = 3 \sin \theta\)

\(2 - 2 \sin^2 \theta = 3 \sin \theta\)

Rearrange to form a quadratic equation:

\(2 \sin^2 \theta + 3 \sin \theta - 2 = 0\)

Factor the quadratic:

\((2 \sin \theta - 1)(\sin \theta + 2) = 0\)

Solving gives \(\sin \theta = \frac{1}{2}\) or \(\sin \theta = -2\). Since \(\sin \theta = -2\) is not possible, we have:

\(\theta = 30^\circ\) or \(150^\circ\).

(ii) Given the smallest positive solution is \(10^\circ\), we have \(n \cdot 10^\circ = 30^\circ\), so \(n = 3\).

For the largest solution, consider \(n \cdot \theta = 150^\circ + 360^\circ k\) for integer \(k\). The largest \(\theta\) in \(0^\circ \leq \theta \leq 360^\circ\) is:

\(3 \cdot \theta = 870^\circ\)

\(\theta = 290^\circ\).

(a) Start with the equation \(3 \tan^2 x - 3 \sin^2 x - 4 = 0\).

Replace \(\tan^2 x\) with \(\frac{\sin^2 x}{\cos^2 x}\) and multiply by \(\cos^2 x\):

\(3 \frac{\sin^2 x}{\cos^2 x} - 3 \sin^2 x - 4 = 0\)

\(3 \sin^2 x - 3 \sin^2 x \cos^2 x - 4 \cos^2 x = 0\)

Replace \(\sin^2 x\) by \(1 - \cos^2 x\):

\(3(1 - \cos^2 x) - 3(1 - \cos^2 x) \cos^2 x - 4 \cos^2 x = 0\)

\(3 - 3 \cos^2 x - 3 \cos^2 x + 3 \cos^4 x - 4 \cos^2 x = 0\)

\(3 \cos^4 x - 10 \cos^2 x + 3 = 0\)

Thus, \(a = 3\), \(b = -10\), \(c = 3\).

(b) Factor the equation:

\((3 \cos^2 x - 1)(\cos^2 x - 3) = 0\)

Solving \(3 \cos^2 x - 1 = 0\):

\(\cos^2 x = \frac{1}{3}\)

\(\cos x = \pm \frac{1}{\sqrt{3}}\)

Solving for \(x\) gives:

\(x = 54.7^\circ, 125.3^\circ\)

(i) Start with the left-hand side (LHS):

\(\tan x + \frac{1}{\tan x} = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\)

Combine the fractions:

\(= \frac{\sin^2 x + \cos^2 x}{\sin x \cos x}\)

Using the identity \(\sin^2 x + \cos^2 x = 1\), we have:

\(= \frac{1}{\sin x \cos x}\)

Thus, the identity is proven.

(ii) Given equation:

\(\frac{2}{\sin x \cos x} = 1 + 3 \tan x\)

Using part (i), substitute \(\frac{2}{\sin x \cos x} = 2 \left( \tan x + \frac{1}{\tan x} \right)\):

\(2 \left( \tan x + \frac{1}{\tan x} \right) = 3 \tan x + 1\)

Multiply through by \(\tan x\):

\(2 \tan^2 x + 2 = 3 \tan x + \tan x\)

Rearrange to form a quadratic equation:

\(2 \tan^2 x + \tan x - 2 = 0\)

Solving the quadratic equation, we find:

\(\tan x = 1\) or \(\tan x = -2\)

For \(\tan x = 1\), \(x = 45^\circ\).

For \(\tan x = -2\), \(x = 116.6^\circ\).

(i) Start with the equation:

\(3 \sin^2 x - 8 \cos x - 7 = 0\).

Using the identity \(\sin^2 x = 1 - \cos^2 x\), substitute:

\(3(1 - \cos^2 x) - 8 \cos x - 7 = 0\).

Simplify to:

\(3 - 3 \cos^2 x - 8 \cos x - 7 = 0\).

\(-3 \cos^2 x - 8 \cos x - 4 = 0\).

Multiply through by -1:

\(3 \cos^2 x + 8 \cos x + 4 = 0\).

Factorize:

\((3 \cos x + 2)(\cos x + 2) = 0\).

Thus, \(\cos x = -\frac{2}{3}\) or \(\cos x = -2\).

Since \(\cos x = -2\) is not possible for real \(x\), we have \(\cos x = -\frac{2}{3}\).

(ii) Substitute \(\cos(\theta + 70^\circ) = -\frac{2}{3}\) into the equation:

\(\theta + 70^\circ = \cos^{-1}(-\frac{2}{3})\).

Calculate \(\theta + 70^\circ\):

\(\theta + 70^\circ = 131.8^\circ\) or \(228.2^\circ\).

For \(0^\circ \leq \theta \leq 180^\circ\), solve:

\(\theta = 131.8^\circ - 70^\circ = 61.8^\circ\).

\(\theta = 228.2^\circ - 70^\circ = 158.2^\circ\).

Thus, \(\theta = 61.8^\circ\) or \(158.2^\circ\).

(i) Start with the given equation:

\(2 \tan^2 \theta \sin^2 \theta = 1\)

Using the identity \(\tan^2 \theta = \frac{\sin^2 \theta}{1 - \sin^2 \theta}\), substitute:

\(2 \left(\frac{\sin^2 \theta}{1 - \sin^2 \theta}\right) \sin^2 \theta = 1\)

Simplify:

\(\frac{2 \sin^4 \theta}{1 - \sin^2 \theta} = 1\)

Multiply both sides by \(1 - \sin^2 \theta\):

\(2 \sin^4 \theta = 1 - \sin^2 \theta\)

Rearrange to get:

\(2 \sin^4 \theta + \sin^2 \theta - 1 = 0\)

(ii) Solve the equation \(2 \sin^4 \theta + \sin^2 \theta - 1 = 0\).

Let \(x = \sin^2 \theta\), then the equation becomes:

\(2x^2 + x - 1 = 0\)

Factorize:

\((2x - 1)(x + 1) = 0\)

So, \(2x - 1 = 0\) or \(x + 1 = 0\).

\(2x - 1 = 0\) gives \(x = \frac{1}{2}\).

\(x + 1 = 0\) gives \(x = -1\), which is not possible since \(x = \sin^2 \theta\) must be non-negative.

Thus, \(\sin^2 \theta = \frac{1}{2}\).

\(\sin \theta = \pm \frac{1}{\sqrt{2}}\).

Solutions for \(\theta\) in the range \(0^\circ \leq \theta \leq 360^\circ\) are:

\(\theta = 45^\circ, 135^\circ, 225^\circ, 315^\circ\).

Start with the equation \(15 \sin^2 x = 13 + \cos x\).

Use the identity \(\sin^2 x = 1 - \cos^2 x\) to rewrite the equation:

\(15(1 - \cos^2 x) = 13 + \cos x\).

Expand and simplify:

\(15 - 15 \cos^2 x = 13 + \cos x\).

Rearrange to form a quadratic equation:

\(15 \cos^2 x + \cos x - 2 = 0\).

Factor the quadratic equation:

\((5 \cos x + 2)(3 \cos x - 1) = 0\).

Solve for \(\cos x\):

\(5 \cos x + 2 = 0\) gives \(\cos x = -\frac{2}{5}\).

\(3 \cos x - 1 = 0\) gives \(\cos x = \frac{1}{3}\).

Find the angles \(x\) for each value of \(\cos x\) within the range \(0^\circ \leq x \leq 180^\circ\):

For \(\cos x = -\frac{2}{5}\), \(x \approx 113.6^\circ\).

For \(\cos x = \frac{1}{3}\), \(x \approx 70.5^\circ\).

(i) Start with the equation \(2 \sin x \tan x + 3 = 0\).

Using \(\tan x = \frac{\sin x}{\cos x}\), rewrite the equation as:

\(2 \sin x \left(\frac{\sin x}{\cos x}\right) + 3 = 0\)

\(\frac{2 \sin^2 x}{\cos x} + 3 = 0\)

Using the identity \(\sin^2 x = 1 - \cos^2 x\), substitute:

\(\frac{2(1 - \cos^2 x)}{\cos x} + 3 = 0\)

Multiply through by \(\cos x\) to clear the fraction:

\(2(1 - \cos^2 x) + 3 \cos x = 0\)

\(2 - 2 \cos^2 x + 3 \cos x = 0\)

Rearrange to get:

\(2 \cos^2 x - 3 \cos x - 2 = 0\)

(ii) Solve the quadratic equation \(2 \cos^2 x - 3 \cos x - 2 = 0\).

Factor the quadratic:

\((2 \cos x + 1)(\cos x - 2) = 0\)

Set each factor to zero:

\(2 \cos x + 1 = 0\) gives \(\cos x = -\frac{1}{2}\)

\(\cos x - 2 = 0\) gives \(\cos x = 2\) (not possible since \(|\cos x| \leq 1\))

For \(\cos x = -\frac{1}{2}\), the solutions in the interval \(0^\circ \leq x \leq 360^\circ\) are:

\(x = 120^\circ\) and \(x = 240^\circ\).

(i) Start with the equation \(2 \tan^2 \theta \cos \theta = 3\).

Replace \(\tan^2 \theta\) with \(\frac{\sin^2 \theta}{\cos^2 \theta}\).

Replace \(\sin^2 \theta\) with \(1 - \cos^2 \theta\).

The equation becomes \(2 \left(\frac{1 - \cos^2 \theta}{\cos^2 \theta}\right) \cos \theta = 3\).

Simplify to get \(2 (1 - \cos^2 \theta) \cos \theta = 3 \cos^2 \theta\).

Rearrange to form \(2 \cos^2 \theta + 3 \cos \theta - 2 = 0\).

(ii) Solve the quadratic equation \(2 \cos^2 \theta + 3 \cos \theta - 2 = 0\).

Factorize to find \(\cos \theta = \frac{1}{2}\) or \(\cos \theta = -2\).

Since \(\cos \theta = -2\) is not possible, solve \(\cos \theta = \frac{1}{2}\).

This gives \(\theta = 60^\circ\) and \(\theta = 300^\circ\) within the range \(0^\circ \leq \theta \leq 360^\circ\).

(i) Start with the equation \(3 \sin x \tan x = 8\).

Use the identity \(\tan x = \frac{\sin x}{\cos x}\), so \(3 \sin x \cdot \frac{\sin x}{\cos x} = 8\).

This simplifies to \(\frac{3 \sin^2 x}{\cos x} = 8\).

Use the identity \(\sin^2 x = 1 - \cos^2 x\), so \(\frac{3(1 - \cos^2 x)}{\cos x} = 8\).

Multiply through by \(\cos x\) to clear the fraction: \(3(1 - \cos^2 x) = 8 \cos x\).

Expand and rearrange: \(3 - 3 \cos^2 x = 8 \cos x\).

Rearrange to form a quadratic: \(3 \cos^2 x + 8 \cos x - 3 = 0\).

(ii) Solve the quadratic equation \(3 \cos^2 x + 8 \cos x - 3 = 0\).

Factor the quadratic: \((3 \cos x - 1)(\cos x + 3) = 0\).

Set each factor to zero: \(3 \cos x - 1 = 0\) or \(\cos x + 3 = 0\).

For \(3 \cos x - 1 = 0\), solve \(\cos x = \frac{1}{3}\).

For \(\cos x + 3 = 0\), \(\cos x = -3\) is not possible since \(\cos x\) must be between -1 and 1.

Find \(x\) for \(\cos x = \frac{1}{3}\) in the range \(0^\circ \leq x \leq 360^\circ\).

\(x = 70.5^\circ\) or \(x = 360^\circ - 70.5^\circ = 289.5^\circ\).

Start with the equation \(3 \sin^2 \theta - 2 \cos \theta - 3 = 0\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to rewrite the equation:

\(3(1 - \cos^2 \theta) - 2 \cos \theta - 3 = 0\)

Simplify to get:

\(3 - 3 \cos^2 \theta - 2 \cos \theta - 3 = 0\)

\(-3 \cos^2 \theta - 2 \cos \theta = 0\)

Factor the equation:

\(\cos \theta (3 \cos \theta + 2) = 0\)

Set each factor to zero:

1. \(\cos \theta = 0\)

\(\theta = 90^\circ\)

2. \(3 \cos \theta + 2 = 0\)

\(\cos \theta = -\frac{2}{3}\)

\(\theta = 131.8^\circ\) (to 1 decimal place)

Thus, the solutions are \(\theta = 90^\circ\) or \(\theta = 131.8^\circ\).

(i) Start with the equation \(\sin^2 \theta + 3 \sin \theta \cos \theta = 4 \cos^2 \theta\).

Divide the entire equation by \(\cos^2 \theta\):

\(\frac{\sin^2 \theta}{\cos^2 \theta} + 3 \frac{\sin \theta \cos \theta}{\cos^2 \theta} = 4\).

This simplifies to \(\tan^2 \theta + 3 \tan \theta = 4\).

Thus, the equation can be written as a quadratic in \(\tan \theta\):

\(\tan^2 \theta + 3 \tan \theta - 4 = 0\).

(ii) Solve the quadratic equation \(\tan^2 \theta + 3 \tan \theta - 4 = 0\).

Using the quadratic formula \(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 3\), \(c = -4\):

\(\tan \theta = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\).

\(\tan \theta = \frac{-3 \pm \sqrt{9 + 16}}{2}\).

\(\tan \theta = \frac{-3 \pm 5}{2}\).

This gives \(\tan \theta = 1\) or \(\tan \theta = -4\).

For \(\tan \theta = 1\), \(\theta = 45^\circ\).

For \(\tan \theta = -4\), \(\theta = 104^\circ\).

Thus, \(\theta = 45^\circ\) or \(\theta = 104^\circ\).

(i) Start with the given equation:

\(4 \sin^4 \theta + 5 = 7 \cos^2 \theta\)

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have \(\cos^2 \theta = 1 - \sin^2 \theta\).

Substitute \(\cos^2 \theta\) in the equation:

\(4 \sin^4 \theta + 5 = 7(1 - \sin^2 \theta)\)

\(4 \sin^4 \theta + 5 = 7 - 7 \sin^2 \theta\)

Let \(x = \sin^2 \theta\), then \(\sin^4 \theta = x^2\).

Substitute \(x\) into the equation:

\(4x^2 + 5 = 7 - 7x\)

Rearrange to form a quadratic equation:

\(4x^2 + 7x - 2 = 0\)

(ii) Solve the quadratic equation \(4x^2 + 7x - 2 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 7\), \(c = -2\):

\(x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}\)

\(x = \frac{-7 \pm \sqrt{49 + 32}}{8}\)

\(x = \frac{-7 \pm \sqrt{81}}{8}\)

\(x = \frac{-7 \pm 9}{8}\)

\(x = \frac{2}{8} = \frac{1}{4} \quad \text{or} \quad x = \frac{-16}{8} = -2\)

Since \(x = \sin^2 \theta\), \(x = -2\) is not valid as \(\sin^2 \theta\) must be between 0 and 1.

Thus, \(\sin^2 \theta = \frac{1}{4}\).

\(\sin \theta = \pm \frac{1}{2}\).

For \(\sin \theta = \frac{1}{2}\), \(\theta = 30^\circ, 150^\circ\).

For \(\sin \theta = -\frac{1}{2}\), \(\theta = 210^\circ, 330^\circ\).

Therefore, \(\theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ\).

(a) (i) Expand \((\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2\sin \theta \cos \theta + \sin^2 \theta = 1\).

This simplifies to \(\cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta = 1\).

Since \(\cos^2 \theta + \sin^2 \theta = 1\), we have \(2\sin \theta \cos \theta = 0\).

This implies \(\sin 2\theta = 0\), leading to \(2\theta = 0, \pi, 2\pi\).

Thus, \(\theta = 0, \frac{1}{2}\pi, \pi\).

(ii) Verify \(\cos \theta + \sin \theta = 1\) for \(\theta = 0\) and \(\theta = \frac{1}{2}\pi\).

For \(\theta = 0\), \(\cos 0 + \sin 0 = 1 + 0 = 1\).

For \(\theta = \frac{1}{2}\pi\), \(\cos \frac{1}{2}\pi + \sin \frac{1}{2}\pi = 0 + 1 = 1\).

(b) Prove the identity:

\(\frac{\sin \theta}{\cos \theta + \sin \theta} + \frac{1 - \cos \theta}{\cos \theta - \sin \theta} = \frac{\sin \theta (\cos \theta - \sin \theta) + (1 - \cos \theta)(\cos \theta + \sin \theta)}{(\cos \theta + \sin \theta)(\cos \theta - \sin \theta)}\).

\(= \frac{\cos \theta \sin \theta - \sin^2 \theta + \cos \theta - \cos^2 \theta + \sin \theta - \cos \theta \sin \theta}{\cos^2 \theta - \sin^2 \theta}\).

\(= \frac{\cos \theta + \sin \theta - 1}{1 - 2 \sin^2 \theta}\).

(c) Solve \(\frac{\cos \theta + \sin \theta - 1}{1 - 2 \sin^2 \theta} = 2(\cos \theta + \sin \theta - 1)\).

Let \(k = \cos \theta + \sin \theta - 1\), then \(\frac{k}{1 - 2 \sin^2 \theta} = 2k\).

\(1 = 2(1 - 2 \sin^2 \theta)\), leading to \(\sin^2 \theta = \frac{1}{4}\).

Thus, \(\sin \theta = \pm \frac{1}{2}\), giving solutions \(\theta = \frac{1}{6}\pi, \frac{5}{6}\pi\).

Including \(\theta = 0\) and \(\theta = \frac{1}{2}\pi\) from part (a)(ii), the solutions are \(0, \frac{1}{6}\pi, \frac{1}{2}\pi, \frac{5}{6}\pi\).

(i) Start with the equation \(3 \tan \theta = 2 \cos \theta\).

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), rewrite as \(3 \frac{\sin \theta}{\cos \theta} = 2 \cos \theta\).

Multiply through by \(\cos \theta\) to get \(3 \sin \theta = 2 \cos^2 \theta\).

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to substitute: \(3 \sin \theta = 2(1 - \sin^2 \theta)\).

Expand and rearrange: \(3 \sin \theta = 2 - 2 \sin^2 \theta\).

Rearrange to form the quadratic: \(2 \sin^2 \theta + 3 \sin \theta - 2 = 0\).

(ii) Solve the quadratic equation \(2 \sin^2 \theta + 3 \sin \theta - 2 = 0\).

Let \(s = \sin \theta\). The equation becomes \(2s^2 + 3s - 2 = 0\).

Factorize to find \(s = 0.5\) or \(s = -2\).

Since \(-1 \leq \sin \theta \leq 1\), only \(s = 0.5\) is valid.

Thus, \(\sin \theta = 0.5\) gives \(\theta = 30^\circ\) or \(150^\circ\) within the range \(0^\circ \leq \theta \leq 360^\circ\).

(i) We start with \(\sin x \tan x = \sin x \cdot \frac{\sin x}{\cos x}\).

This simplifies to \(\frac{\sin^2 x}{\cos x}\).

Using the identity \(\sin^2 x = 1 - \cos^2 x\), we have:

\(\frac{1 - \cos^2 x}{\cos x}\).

Thus, \(\sin x \tan x = \frac{1 - \cos^2 x}{\cos x}\).

(ii) Given \(2 \sin x \tan x = 3\), substitute \(\sin x \tan x = \frac{1 - \cos^2 x}{\cos x}\):

\(2 \cdot \frac{1 - \cos^2 x}{\cos x} = 3\).

Multiply through by \(\cos x\):

\(2(1 - \cos^2 x) = 3 \cos x\).

Expand and rearrange:

\(2 - 2\cos^2 x = 3\cos x\).

\(2\cos^2 x + 3\cos x - 2 = 0\).

Let \(c = \cos x\), then:

\(2c^2 + 3c - 2 = 0\).

Using the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = 3, c = -2\):

\(c = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}\).

\(c = \frac{-3 \pm \sqrt{9 + 16}}{4}\).

\(c = \frac{-3 \pm 5}{4}\).

\(c = \frac{2}{4} = 0.5\) or \(c = \frac{-8}{4} = -2\).

Since \(\cos x\) must be between -1 and 1, \(\cos x = 0.5\).

Thus, \(x = 60^\circ\) or \(x = 300^\circ\).

Start with the equation \(\tan \theta \sin \theta = 1\).

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have:

\(\frac{\sin^2 \theta}{\cos \theta} = 1\).

Multiply both sides by \(\cos \theta\):

\(\sin^2 \theta = \cos \theta\).

Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\), substitute:

\(1 - \cos^2 \theta = \cos \theta\).

Rearrange to form a quadratic equation:

\(\cos^2 \theta + \cos \theta - 1 = 0\).

Use the quadratic formula \(\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1, b = 1, c = -1\):

\(\cos \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}\).

\(\cos \theta = \frac{-1 \pm \sqrt{5}}{2}\).

Calculate the possible values for \(\cos \theta\):

\(\cos \theta = \frac{-1 + \sqrt{5}}{2} \approx 0.618\).

Find \(\theta\) using \(\cos^{-1}(0.618)\):

\(\theta \approx 51.8^\circ\) and \(\theta \approx 360^\circ - 51.8^\circ = 308.2^\circ\).

Start with the equation \(8 \sin^2 \theta + 6 \cos \theta + 1 = 0\).

Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\), substitute to get:

\(8(1 - \cos^2 \theta) + 6 \cos \theta + 1 = 0\).

Simplify to:

\(8 - 8 \cos^2 \theta + 6 \cos \theta + 1 = 0\).

Combine like terms:

\(-8 \cos^2 \theta + 6 \cos \theta + 9 = 0\).

Rearrange to:

\(8 \cos^2 \theta - 6 \cos \theta - 9 = 0\).

Factor the quadratic equation:

\((4 \cos \theta + 3)(2 \cos \theta - 3) = 0\).

Set each factor to zero:

\(4 \cos \theta + 3 = 0\) or \(2 \cos \theta - 3 = 0\).

Solve for \(\cos \theta\):

\(\cos \theta = -\frac{3}{4}\).

Find \(\theta\) using the inverse cosine function:

\(\theta = \cos^{-1}(-0.75)\).

Within the range \(0^\circ < \theta < 180^\circ\), \(\theta = 138.6^\circ\).

(a) Start with the left-hand side: \(\frac{\sin \theta}{\sin \theta + \cos \theta} + \frac{\cos \theta}{\sin \theta - \cos \theta}\).

Combine the fractions: \(\frac{\sin \theta (\sin \theta - \cos \theta) + \cos \theta (\sin \theta + \cos \theta)}{(\sin \theta + \cos \theta)(\sin \theta - \cos \theta)}\).

Simplify the numerator: \(\sin^2 \theta - \sin \theta \cos \theta + \cos \theta \sin \theta + \cos^2 \theta = \sin^2 \theta + \cos^2 \theta = 1\).

The denominator becomes \(\sin^2 \theta - \cos^2 \theta\).

Thus, the expression simplifies to \(\frac{1}{\sin^2 \theta - \cos^2 \theta}\).

Using the identity \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\), rewrite as \(\frac{1}{\frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta}} = \frac{\cos^2 \theta}{\sin^2 \theta - \cos^2 \theta}\).

Recognize \(\sin^2 \theta - \cos^2 \theta = (\tan^2 \theta + 1) - 2\), so \(\frac{\cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} = \frac{\tan^2 \theta + 1}{\tan^2 \theta - 1}\).

Thus, the identity is proven.

(b) Set \(\frac{\tan^2 \theta + 1}{\tan^2 \theta - 1} = 2\).

Cross-multiply to get \(\tan^2 \theta + 1 = 2(\tan^2 \theta - 1)\).

Simplify: \(\tan^2 \theta + 1 = 2\tan^2 \theta - 2\).

Rearrange: \(\tan^2 \theta = 3\).

Thus, \(\tan \theta = \pm \sqrt{3}\).

For \(\tan \theta = \sqrt{3}\), \(\theta = \frac{\pi}{3}\).

For \(\tan \theta = -\sqrt{3}\), \(\theta = \frac{2\pi}{3}\).

Therefore, the solutions are \(\theta = \frac{1}{3} \pi, \frac{2}{3} \pi\).

First, factor the quadratic equation:

\(8 \cos^2 \theta - 10 \cos \theta + 2 = 0\)

can be factored as \(2(4 \cos \theta - 1)(\cos \theta - 1) = 0\).

This gives two equations:

1. \(4 \cos \theta - 1 = 0\)

2. \(\cos \theta - 1 = 0\)

Solving the first equation:

\(4 \cos \theta = 1\)

\(\cos \theta = \frac{1}{4}\)

\(\theta = 75.5^\circ\) (using a calculator for inverse cosine)

Solving the second equation:

\(\cos \theta = 1\)

\(\theta = 0^\circ\)

Thus, the solutions are \(\theta = 0^\circ\) and \(\theta = 75.5^\circ\).