Start with the equation:

\(4 \sin \theta + \tan \theta = 0\)

Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):

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

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

\(4 \sin \theta \cos \theta + \sin \theta = 0\)

Factor out \(\sin \theta\):

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

This gives two cases:

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

2. \(4 \cos \theta + 1 = 0\)

For \(\sin \theta = 0\), \(\theta = 0^\circ, 180^\circ\), but these are outside the given range.

For \(4 \cos \theta + 1 = 0\):

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

Using the inverse cosine function, \(\theta \approx 104.5^\circ\) within the range \(0^\circ < \theta < 180^\circ\).

(a) Start with the left-hand side: \(\frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta}\).

Combine the fractions: \(\frac{(1 + \sin \theta)^2 + \cos^2 \theta}{\cos \theta (1 + \sin \theta)}\).

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

Using \(\sin^2 \theta + \cos^2 \theta = 1\), the numerator becomes \(2 + 2\sin \theta\).

Thus, \(\frac{2 + 2\sin \theta}{\cos \theta (1 + \sin \theta)} = \frac{2}{\cos \theta}\), proving the identity.

(b) Given \(\frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta} = \frac{3}{\sin \theta}\), use the identity from part (a): \(\frac{2}{\cos \theta} = \frac{3}{\sin \theta}\).

Cross-multiply to get \(2\sin \theta = 3\cos \theta\).

Divide both sides by \(\cos \theta\): \(\tan \theta = 1.5\).

Find \(\theta\) using \(\tan^{-1}(1.5)\): \(\theta = 0.983\).

Since \(\tan \theta\) is periodic with period \(\pi\), the second solution is \(\theta = 0.983 + \pi = 4.12\).

Start with the equation:

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

Multiply both sides by the denominator:

\(\tan \theta + 3 \sin \theta + 2 = 2(\tan \theta - 3 \sin \theta + 1)\)

Expand and simplify:

\(\tan \theta + 3 \sin \theta + 2 = 2 \tan \theta - 6 \sin \theta + 2\)

Rearrange terms:

\(2 \tan \theta - \tan \theta - 6 \sin \theta - 3 \sin \theta + 2 - 2 = 0\)

\(\tan \theta - 9 \sin \theta = 0\)

Multiply by \(\cos \theta\):

\(\sin \theta - 9 \sin \theta \cos \theta = 0\)

Factorize:

\(\sin \theta (1 - 9 \cos \theta) = 0\)

Set each factor to zero:

\(\sin \theta = 0\) or \(1 - 9 \cos \theta = 0\)

For \(\sin \theta = 0\), \(\theta = 0^\circ\).

For \(1 - 9 \cos \theta = 0\), \(\cos \theta = \frac{1}{9}\).

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

\(\theta \approx 83.6^\circ\).

Thus, \(\theta = 0^\circ\) or \(83.6^\circ\) (only answers in the given range).

Start with the equation \(3 \tan(2x + 1) = 1\).

Divide both sides by 3: \(\tan(2x + 1) = \frac{1}{3}\).

Take the inverse tangent: \(2x + 1 = \tan^{-1}\left(\frac{1}{3}\right)\).

Using a calculator, \(\tan^{-1}\left(\frac{1}{3}\right) \approx 0.322\) radians.

Thus, \(2x + 1 = 0.322 + k\pi\) or \(2x + 1 = -0.339 + k\pi\) for integer \(k\).

Solving for \(x\):

\(2x = 0.322 - 1 + k\pi\) or \(2x = -0.339 - 1 + k\pi\).

\(x = \frac{-0.678 + k\pi}{2}\) or \(x = \frac{-1.339 + k\pi}{2}\).

For the smallest positive \(x\), set \(k = 1\):

\(x = \frac{-0.678 + \pi}{2} \approx 1.23\).

For the next smallest \(x\), set \(k = 2\):

\(x = \frac{-1.339 + 2\pi}{2} \approx 2.80\).

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

\(\left( \frac{1}{\cos x} - \tan x \right)^2 = \left( \frac{1}{\cos x} - \frac{\sin x}{\cos x} \right)^2 = \left( \frac{1 - \sin x}{\cos x} \right)^2\)

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

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

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

\(= \frac{1 - \sin x}{1 + \sin x}\)

This matches the right-hand side (RHS), proving the identity.

(ii) Using the result from part (i), we have:

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

Cross-multiply to get:

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

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

\(2 = 4\sin 2x\)

\(\sin 2x = \frac{1}{2}\)

Solving \(\sin 2x = \frac{1}{2}\), we find:

\(2x = \frac{\pi}{6}\) or \(2x = \frac{5\pi}{6}\)

Thus, \(x = \frac{\pi}{12}\) or \(x = \frac{5\pi}{12}\).

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

Multiply numerator and denominator by \(\cos^2 \theta\):

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

Using \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to \(\sin^2 \theta - \cos^2 \theta = \sin^2 \theta - (1 - \sin^2 \theta) = 2\sin^2 \theta - 1\).

Thus, \(a = 2\) and \(b = -1\).

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

\(2\sin^2 \theta = \frac{1}{4} + 1 = \frac{5}{4}\).

\(\sin^2 \theta = \frac{5}{8}\).

\(\sin \theta = \pm \sqrt{\frac{5}{8}} \approx \pm 0.7906\).

Since \(-90^\circ \leq \theta \leq 0^\circ\), \(\theta = -52.2^\circ\).

(i) To prove the identity \((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \equiv \sin^3 \theta + \cos^3 \theta\):

Expand the left-hand side (LHS):

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

Rearrange terms:

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

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

\(= \sin^3 \theta + \cos^3 \theta\)

Thus, the identity is proven.

(ii) Solve \((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) = 3 \cos^3 \theta\):

From part (i), \(\sin^3 \theta + \cos^3 \theta = 3 \cos^3 \theta\)

Rearrange to get:

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

Divide both sides by \(\cos^3 \theta\):

\(\tan^3 \theta = 2\)

Taking the cube root gives:

\(\tan \theta = \sqrt[3]{2}\)

Calculate \(\theta\) using \(\tan^{-1}(\sqrt[3]{2})\):

\(\theta = 51.6^\circ\) or \(231.6^\circ\) (considering the periodicity of tangent).

(i) Start with the given equation:

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

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

\(\frac{2 \sin \theta + \cos \theta}{\sin \theta + \cos \theta} = 2 \cdot \frac{\sin \theta}{\cos \theta}\)

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

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

Expand both sides:

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

Subtract \(2 \sin \theta \cos \theta\) from both sides:

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

(ii) From \(\cos^2 \theta = 2 \sin^2 \theta\), use the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

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

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

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

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

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

For \(0^\circ < \theta < 180^\circ\), \(\theta = 35.3^\circ\) or \(144.7^\circ\).

(i) Start with the left-hand side: \(\left( \frac{1}{\cos \theta} - \tan \theta \right)^2\).

Rewrite \(\tan \theta\) as \(\frac{\sin \theta}{\cos \theta}\), so the expression becomes \(\left( \frac{1 - \sin \theta}{\cos \theta} \right)^2\).

This simplifies to \(\frac{(1 - \sin \theta)^2}{\cos^2 \theta}\).

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), the expression becomes \(\frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta}\).

Factor the denominator: \(1 - \sin^2 \theta = (1 - \sin \theta)(1 + \sin \theta)\).

Cancel \(1 - \sin \theta\) from the numerator and denominator to get \(\frac{1 - \sin \theta}{1 + \sin \theta}\), which matches the right-hand side.

(ii) From part (i), we have \(\left( \frac{1}{\cos \theta} - \tan \theta \right)^2 = \frac{1 - \sin \theta}{1 + \sin \theta}\).

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

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

Simplify to \(2 - 2\sin \theta = 1 + \sin \theta\).

Rearrange to \(2 - 1 = 2\sin \theta + \sin \theta\).

Thus, \(1 = 3\sin \theta\) or \(\sin \theta = \frac{1}{3}\).

Find \(\theta\) using \(\sin^{-1} \left( \frac{1}{3} \right)\), which gives \(\theta = 19.5^\circ\) or \(\theta = 160.5^\circ\) within the range \(0^\circ \leq \theta \leq 360^\circ\).

(i) Start with the left-hand side: \(\frac{1 + \cos \theta}{\sin \theta} + \frac{\sin \theta}{1 + \cos \theta}\).

Combine the fractions: \(\frac{(1 + \cos \theta)^2 + \sin^2 \theta}{\sin \theta (1 + \cos \theta)}\).

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

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

So, \(\frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta}\).

This proves the identity.

(ii) Use the result from part (i): \(\frac{2}{\sin \theta} = \frac{3}{\cos \theta}\).

Cross-multiply: \(2\cos \theta = 3\sin \theta\).

Divide both sides by \(\cos \theta \sin \theta\): \(\frac{2}{\sin \theta} = \frac{3}{\cos \theta} \rightarrow t = \frac{2}{3}\).

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

Find \(\theta\) using inverse tangent: \(\theta = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.7^\circ\).

Since \(\tan \theta\) is positive in the first and third quadrants, the solutions are \(\theta = 33.7^\circ\) and \(\theta = 180^\circ + 33.7^\circ = 213.7^\circ\).

Start with the given equation:

\(6 \sin^2 x - 5 \cos^2 x = 2 \sin^2 x + \cos^2 x\)

Rearrange terms:

\(6 \sin^2 x - 2 \sin^2 x = 5 \cos^2 x + \cos^2 x\)

\(4 \sin^2 x = 6 \cos^2 x\)

Divide both sides by \(\cos^2 x\):

\(\frac{4 \sin^2 x}{\cos^2 x} = 6\)

\(4 \tan^2 x = 6\)

\(\tan^2 x = \frac{6}{4} = 1.5\)

\(\tan x = \pm \sqrt{1.5}\)

\(\tan x = \pm 1.225\)

Find the angles \(x\) in the range \(0^\circ \leq x \leq 360^\circ\):

\(x = 50.8^\circ, 129.2^\circ, 230.8^\circ, 309.2^\circ\)

Start with the equation:

\(4 \sin\left(\frac{1}{2}x - 30^\circ\right) = 2\sqrt{2}\)

Divide both sides by 4:

\(\sin\left(\frac{1}{2}x - 30^\circ\right) = \frac{\sqrt{2}}{2}\)

Recognize that \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = 45^\circ\) or \(135^\circ\).

Set up the equations:

\(\frac{1}{2}x - 30^\circ = 45^\circ\)

\(\frac{1}{2}x - 30^\circ = 135^\circ\)

Solve each equation for \(x\):

For \(45^\circ\):

\(\frac{1}{2}x = 45^\circ + 30^\circ\)

\(\frac{1}{2}x = 75^\circ\)

\(x = 150^\circ\)

For \(135^\circ\):

\(\frac{1}{2}x = 135^\circ + 30^\circ\)

\(\frac{1}{2}x = 165^\circ\)

\(x = 330^\circ\)

Thus, the solutions are \(x = 150^\circ\) and \(x = 330^\circ\).

(i) Start with the equation:

\(\sin 2x + 3 \cos 2x = 3(\sin 2x - \cos 2x)\)

Expand the right side:

\(\sin 2x + 3 \cos 2x = 3\sin 2x - 3\cos 2x\)

Rearrange terms:

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

\(-2\sin 2x + 6\cos 2x = 0\)

Divide by \(\cos 2x\):

\(-2\tan 2x + 6 = 0\)

\(\tan 2x = 3\)

Thus, \(k = 3\).

(ii) Solve \(\tan 2x = 3\) for \(-90^\circ \leq x \leq 90^\circ\):

\(2x = \tan^{-1}(3)\)

\(2x = 71.6^\circ\) or \(2x = -108.4^\circ\)

Divide by 2:

\(x = 35.8^\circ\) or \(x = -54.2^\circ\)

(i) Start by combining the fractions on the left-hand side:

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

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

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

\(= \frac{4}{\sin \theta \tan \theta}\)

Thus, the identity is proven.

(ii) Using the result from part (i), substitute into the equation:

\(\sin \theta \times \frac{4}{\sin \theta \tan \theta} = 3\)

\(\frac{4}{\tan \theta} = 3\)

\(\tan \theta = \frac{4}{3}\)

Solving for \(\theta\), we find:

\(\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.1^\circ\)

Considering the periodicity of the tangent function, the other solution in the range is:

\(\theta = 180^\circ + 53.1^\circ = 233.1^\circ\)

(i) Start with \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 = \left( \frac{s - c}{s} \right)^2\) where \(s = \sin x\) and \(c = \cos x\).

\(= \frac{(1-c)^2}{s^2} = \frac{(1-c)^2}{1-c^2}\) using \(s^2 = 1-c^2\).

\(= \frac{(1-c)(1-c)}{(1-c)(1+c)} = \frac{1-c}{1+c}\).

Thus, \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 \equiv \frac{1 - \cos x}{1 + \cos x}\).

(ii) Given \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 = \frac{2}{5}\), equate to \(\frac{1-c}{1+c} = \frac{2}{5}\).

Solving for \(c\), \(\frac{1-c}{1+c} = \frac{2}{5}\) gives \(5(1-c) = 2(1+c)\).

\(5 - 5c = 2 + 2c\) leads to \(3 = 7c\) or \(c = \frac{3}{7}\).

Thus, \(\cos x = \frac{3}{7}\).

Solving for \(x\), \(x = \cos^{-1} \left( \frac{3}{7} \right) \approx 1.13\) or \(x = 2\pi - 1.13 \approx 5.16\).

(i) Start with the equation \(3 \sin \theta = \cos \theta\). Divide both sides by \(\cos \theta\) to get \(3 \tan \theta = 1\). Therefore, \(\tan \theta = \frac{1}{3}\). Solving for \(\theta\) in the range \(0^\circ < \theta < 180^\circ\), we find \(\theta = 18.4^\circ\).

(ii) Given \(3 \sin^2 2x = \cos^2 2x\), rewrite it as \(\frac{\sin^2 2x}{\cos^2 2x} = \frac{1}{3}\). This implies \(\tan^2 2x = \frac{1}{3}\), so \(\tan 2x = \pm \frac{1}{\sqrt{3}}\). The solutions for \(2x\) are \(30^\circ, 150^\circ, 210^\circ, 330^\circ\). Dividing by 2, we get \(x = 15^\circ, 75^\circ, 105^\circ, 165^\circ\).

(i) We start with the identity \(\sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)\).

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

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

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

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

This completes the proof.

(ii) Given \(\sin^4 \theta - \cos^4 \theta = \frac{1}{2}\), we use the result from part (i):

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

Solving for \(\sin^2 \theta\), we have:

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

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

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

Thus, \(\theta = 60^\circ, 120^\circ, 240^\circ, 300^\circ\).

First, multiply both sides by \(2 + \cos \theta\) to eliminate the fraction:

\(13 \sin^2 \theta + \cos \theta (2 + \cos \theta) = 2(2 + \cos \theta)\)

Simplify the equation:

\(13 \sin^2 \theta + 2 \cos \theta + \cos^2 \theta = 4 + 2 \cos \theta\)

Rearrange terms:

\(13 \sin^2 \theta + \cos^2 \theta = 4\)

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

\(13(1 - \cos^2 \theta) + \cos^2 \theta = 4\)

Simplify:

\(13 - 13 \cos^2 \theta + \cos^2 \theta = 4\)

\(13 - 12 \cos^2 \theta = 4\)

\(12 \cos^2 \theta = 9\)

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

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

Thus, \(\theta = 30^\circ, 150^\circ\).

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

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

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

Since \(\sin^2 x + \cos^2 x = 1\), this simplifies to:

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

Thus, the identity is proven.

(ii) Given \(\frac{\tan x + 1}{\sin x \tan x + \cos x} = 3 \sin x - 2 \cos x\), use the proven identity:

\(\sin x + \cos x = 3 \sin x - 2 \cos x\)

Rearrange to find:

\(\cos x + 2 \cos x = 3 \sin x - \sin x\)

\(3 \cos x = 2 \sin x\)

\(\tan x = \frac{2}{3}\)

Find \(x\) using \(\tan^{-1} \left( \frac{2}{3} \right)\):

\(x \approx 0.983\)

Also consider the periodicity of tangent:

\(x \approx 0.983 + \pi \approx 4.12 \text{ or } 4.13\)

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

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

Combine the fractions:

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

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

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

Factor out \(\sin \theta\):

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

Cancel \(1 + \sin \theta\):

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

Thus, the identity is proven.

(ii) Solve the equation:

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

Using part (i), replace the expression with \(\tan \theta\):

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

\(\tan \theta = -2\)

Find \(\theta\) such that \(\tan \theta = -2\) in the range \(0^\circ \leq \theta \leq 360^\circ\):

\(\theta = 116.6^\circ\) or \(296.6^\circ\)

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

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

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

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

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

Thus, the identity is proven.

(ii) Solve \(\frac{1}{\tan \theta} = 4 \tan \theta\).

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

\(\tan \theta = \pm \frac{1}{2}\)

For \(\tan \theta = \frac{1}{2}\), \(\theta = 26.6^\circ\).

For \(\tan \theta = -\frac{1}{2}\), \(\theta = 153.4^\circ\).

(i) Start with the expression:

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

Combine the fractions:

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

Expand the numerator:

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

Which simplifies to:

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

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

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

(ii) Solve the equation:

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

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

Using \(\sin^2 \theta + \cos^2 \theta = 1\), let \(s = \sin \theta\) and \(c = \cos \theta\):

\(s^2 - c^2 = \frac{1}{3}\)

\(s^2 + c^2 = 1\)

Subtract the equations:

\(2c^2 = \frac{2}{3}\)

\(c^2 = \frac{1}{3}\)

\(c = \pm \frac{1}{\sqrt{3}}\)

\(s^2 = \frac{2}{3}\)

\(s = \pm \sqrt{\frac{2}{3}}\)

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

Solutions for \(\theta\) are:

\(\theta = 54.7^\circ, 125.3^\circ, 234.7^\circ, 305.3^\circ\)

Start with the equation:

\(6 \cos \theta \tan \theta - 3 \cos \theta + 4 \tan \theta - 2 = 0\)

Factorise by grouping:

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

Factor out \((2 \tan \theta - 1)\):

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

Set each factor to zero:

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

Solve for \(\theta\):

\(\tan \theta = \frac{1}{2}\)

\(\theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.6^\circ\)

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

\(\theta = \cos^{-1}\left(-\frac{2}{3}\right) \approx 131.8^\circ\)

Thus, the solutions are \(\theta = 26.6^\circ\) and \(\theta = 131.8^\circ\).

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

Rearrange to \(\tan 2x = -3\).

Find \(2x\) using the inverse tangent: \(2x = 180^\circ - 71.6^\circ\) or \(360^\circ - 71.6^\circ\).

This gives \(2x = 108.4^\circ\) or \(2x = 288.4^\circ\).

Divide by 2 to find \(x\): \(x = 54.2^\circ\) or \(x = 144.2^\circ\).

Also consider the periodicity of tangent: \(x = 234.2^\circ\) and \(x = 324.2^\circ\).

(ii) The range \(0^\circ \leq x \leq 1080^\circ\) is three times the range \(0^\circ \leq x \leq 360^\circ\), so there are 3 times the number of solutions: 12 answers.

Given the equation \(\sin 2x = 2 \cos 2x\).

Divide both sides by \(\cos 2x\) (assuming \(\cos 2x \neq 0\)) to get:

\(\tan 2x = 2\).

Find the general solution for \(2x\):

\(2x = \tan^{-1}(2) + k \cdot 180^\circ\), where \(k\) is an integer.

Calculate \(\tan^{-1}(2) \approx 63.4^\circ\).

Thus, \(2x = 63.4^\circ\) or \(2x = 243.4^\circ\) (adding \(180^\circ\)).

Divide by 2 to find \(x\):

\(x = 31.7^\circ\) or \(x = 121.7^\circ\).

Both solutions are within the given range \(0^\circ \leq x \leq 180^\circ\).

(i) Start with \(\left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right)^2\).

Rewrite \(\frac{1}{\tan \theta}\) as \(\frac{\cos \theta}{\sin \theta}\), so:

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

This becomes \(\frac{(1 - \cos \theta)^2}{\sin^2 \theta}\).

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

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

(ii) Use the result from part (i):

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

Cross-multiply to get:

\(5(1 - \cos \theta) = 2(1 + \cos \theta)\).

Simplify to find \(\cos \theta\):

\(5 - 5\cos \theta = 2 + 2\cos \theta\).

\(3 = 7\cos \theta\).

\(\cos \theta = \frac{3}{7}\).

Find \(\theta\) using inverse cosine:

\(\theta = 64.6^\circ\) or \(295.4^\circ\).

(i) Start with the left-hand side:

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

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

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

Factor the numerator:

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

Cancel \((1 - \sin \theta)\):

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

Thus, the identity is proven.

(ii) Solve \(\frac{\cos \theta}{\tan \theta (1 - \sin \theta)} = 4\):

From part (i), \(1 + \frac{1}{\sin \theta} = 4\).

\(\frac{1}{\sin \theta} = 3\)

\(\sin \theta = \frac{1}{3}\)

Find \(\theta\) in the range \(0^\circ \leq \theta \leq 360^\circ\):

\(\theta = \arcsin \left( \frac{1}{3} \right) \approx 19.5^\circ\)

Also, \(\theta = 180^\circ - 19.5^\circ = 160.5^\circ\).

Thus, \(\theta = 19.5^\circ, 160.5^\circ\).

(i) Start with the left-hand side: \(\frac{\sin x \tan x}{1 - \cos x} = \frac{\sin x \cdot \frac{\sin x}{\cos x}}{1 - \cos x} = \frac{\sin^2 x}{\cos x (1 - \cos x)}\).

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

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

Cancel \(1 - \cos x\) from numerator and denominator:

\(\frac{1 + \cos x}{\cos x} = 1 + \frac{1}{\cos x}\).

Thus, the identity is proven.

(ii) From part (i), \(\frac{1}{\cos x} + 1 + 2 = 0\) simplifies to \(\frac{1}{\cos x} = -3\).

Therefore, \(\cos x = -\frac{1}{3}\).

Solving \(\cos x = -\frac{1}{3}\) for \(0^\circ \leq x \leq 360^\circ\), we find:

\(x = 109.5^\circ\) or \(x = 250.5^\circ\).

(i) Start with the equation:

\(3(2 \sin x - \cos x) = 2(\sin x - 3 \cos x)\)

Expand both sides:

\(6 \sin x - 3 \cos x = 2 \sin x - 6 \cos x\)

Rearrange to collect like terms:

\(6 \sin x - 2 \sin x = -6 \cos x + 3 \cos x\)

\(4 \sin x = -3 \cos x\)

Divide both sides by \(\cos x\):

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

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

(ii) From \(\tan x = -\frac{3}{4}\), find the general solution:

\(x = \tan^{-1}\left(-\frac{3}{4}\right)\)

\(x \approx -36.9^\circ\)

Adjust for the range \(0^\circ \leq x \leq 360^\circ\):

\(x = 180^\circ - 36.9^\circ = 143.1^\circ\)

\(x = 360^\circ - 36.9^\circ = 323.1^\circ\)

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

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

Use identities: \(\sin^2 x = 1 - \cos^2 x\) and \(\cos^2 x = 1 - \sin^2 x\).

Substitute: \(\sin x \cos^2 x = \sin x (1 - \sin^2 x) = \sin x - \sin^3 x\) and \(\cos x \sin^2 x = \cos x (1 - \cos^2 x) = \cos x - \cos^3 x\).

Combine: \(\sin x + \cos x - (\sin x - \sin^3 x) - (\cos x - \cos^3 x) = \sin^3 x + \cos^3 x\).

Thus, the identity is proven.

(ii) Given equation: \((\sin x + \cos x)(1 - \sin x \cos x) = 9 \sin^3 x\).

Use part (i): \(\sin^3 x + \cos^3 x = 9 \sin^3 x\).

Rearrange: \(\cos^3 x = 8 \sin^3 x\).

Divide by \(\cos^3 x\): \(\tan^3 x = \frac{1}{8}\).

Take cube root: \(\tan x = \frac{1}{2}\).

Find solutions: \(x = \tan^{-1}(\frac{1}{2}) \approx 26.6^\circ\).

Use periodicity: \(x = 26.6^\circ + 180^\circ = 206.6^\circ\).

Thus, the solutions are \(x = 26.6^\circ\) and \(x = 206.6^\circ\).

Start with the equation \(3 \tan(2x + 15^\circ) = 4\).

Divide both sides by 3 to isolate the tangent function:

\(\tan(2x + 15^\circ) = \frac{4}{3}\).

Find the angle whose tangent is \(\frac{4}{3}\):

\(2x + 15^\circ = \tan^{-1}\left(\frac{4}{3}\right)\).

Using a calculator, \(\tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ\).

Since tangent has a period of \(180^\circ\), the general solution is:

\(2x + 15^\circ = 53.13^\circ + 180^\circ k\) or \(2x + 15^\circ = 233.13^\circ + 180^\circ k\), where \(k\) is an integer.

For \(k = 0\), solve for \(x\):

\(2x + 15^\circ = 53.13^\circ\)

\(2x = 53.13^\circ - 15^\circ\)

\(2x = 38.13^\circ\)

\(x = 19.065^\circ\)

Rounding to one decimal place, \(x = 19.1^\circ\).

For the second solution:

\(2x + 15^\circ = 233.13^\circ\)

\(2x = 233.13^\circ - 15^\circ\)

\(2x = 218.13^\circ\)

\(x = 109.065^\circ\)

Rounding to one decimal place, \(x = 109.1^\circ\).

Thus, the solutions are \(x = 19.1^\circ\) or \(x = 109.1^\circ\).

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

Rearrange to find \(\tan 2x\):

\(\tan 2x = -3\).

Find the angle \(2x\) using the inverse tangent function:

\(2x = 180^\circ - 71.6^\circ\) or \(2x = 360^\circ - 71.6^\circ\).

This gives \(2x = 108.4^\circ\) or \(2x = 288.4^\circ\).

Divide by 2 to solve for \(x\):

\(x = 54.2^\circ\) or \(x = 144.2^\circ\).

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

Let \(s = \sin \theta\) and \(c = \cos \theta\).

Then the equation becomes \(s + c = 2s - 2c\).

Rearrange to get \(s + c = 2s - 2c \Rightarrow s + c = 2s - 2c \Rightarrow s = 3c\).

Divide both sides by \(c\) (assuming \(c \neq 0\)) to get \(\frac{s}{c} = 3\).

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

(ii) From \(\tan \theta = 3\), find \(\theta\) in the range \(0^\circ \leq \theta \leq 360^\circ\).

The principal value is \(\theta = \tan^{-1}(3) \approx 71.6^\circ\).

The other solution in the range is \(\theta = 180^\circ + 71.6^\circ = 251.6^\circ\).

Thus, \(\theta = 71.6^\circ\) or \(251.6^\circ\).

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

Multiply numerator and denominator by \(\cos x\):

\(\frac{\sin x + \sin x \cos x}{\sin x - \sin x \cos x} = k\).

Rearrange to \(\frac{1 + \cos x}{1 - \cos x} = k\).

(b) From \(\frac{1 + \cos x}{1 - \cos x} = k\), cross-multiply to get:

\(k - k \cos x = 1 + \cos x\).

Rearrange to \(k - 1 = (k + 1) \cos x\).

Thus, \(\cos x = \frac{k - 1}{k + 1}\).

(c) Substitute \(k = 4\) into \(\cos x = \frac{k - 1}{k + 1}\):

\(\cos x = \frac{4 - 1}{4 + 1} = \frac{3}{5}\).

Find \(x\) such that \(\cos x = \frac{3}{5}\) within the range \(-\pi < x < \pi\).

The solutions are \(x = \pm 0.927\) (only solutions in the given range).

Given the equation \(\sin 3x + 2 \cos 3x = 0\).

We can rewrite this as \(\tan 3x = -2\) by dividing both sides by \(\cos 3x\).

Solving \(\tan 3x = -2\), we find the principal value:

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

Since \(\tan\) is periodic with period \(180^\circ\), the general solution for \(3x\) is:

\(3x = -63.4^\circ + 180^\circ n\), where \(n\) is an integer.

Solving for \(x\), we have:

\(x = \frac{-63.4^\circ + 180^\circ n}{3}\).

We need \(0^\circ \leq x \leq 180^\circ\), so:

\(0^\circ \leq \frac{-63.4^\circ + 180^\circ n}{3} \leq 180^\circ\).

Solving for \(n\), we find the valid values:

For \(n = 1\), \(x = \frac{116.6^\circ}{3} = 38.9^\circ\).

For \(n = 2\), \(x = \frac{296.6^\circ}{3} = 98.9^\circ\).

For \(n = 3\), \(x = \frac{476.6^\circ}{3} = 158.9^\circ\).

Thus, the solutions are \(x = 38.9^\circ, 98.9^\circ, 158.9^\circ\).

(a) To prove the identity:

Start with \(\frac{1 + \sin x}{1 - \sin x} - \frac{1 - \sin x}{1 + \sin x}\).

Use a common denominator: \((1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x\).

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

Thus, \(\frac{4\sin x}{\cos^2 x} = \frac{4\sin x}{\cos x \cdot \cos x} = \frac{4\tan x}{\cos x}\).

Identity is proven.

(b) Solve \(\frac{4\tan x}{\cos x} = 8\tan x\).

Divide both sides by \(\tan x\) (assuming \(\tan x \neq 0\)): \(\frac{4}{\cos x} = 8\).

\(\cos x = \frac{1}{2}\).

\(x = \frac{\pi}{3}\) within the given range.

Also, consider \(\tan x = 0\) which gives \(x = 0\).

Thus, solutions are \(x = 0\) or \(x = \frac{\pi}{3}\).

Start with the equation:

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

Cross-multiply to get:

\(\tan \theta + 2 \sin \theta = 3(\tan \theta - 2 \sin \theta)\)

Expand the right side:

\(\tan \theta + 2 \sin \theta = 3 \tan \theta - 6 \sin \theta\)

Rearrange terms:

\(2 \tan \theta - 8 \sin \theta = 0\)

Factor out \(2 \sin \theta\):

\(2 \sin \theta (\tan \theta - 4 \cos \theta) = 0\)

Since \(\sin \theta = 0\) is not valid in the given range, solve:

\(\tan \theta = 4 \cos \theta\)

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

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

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

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

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

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

Rearrange to form a quadratic equation:

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

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

\(\sin \theta = \frac{-1 \pm \sqrt{1 + 64}}{8}\)

\(\sin \theta = \frac{-1 \pm 9}{8}\)

\(\sin \theta = 1\) or \(\sin \theta = -\frac{5}{4}\)

Only \(\sin \theta = 1\) is valid, leading to \(\theta = 90^\circ\), but this is not in the solution range.

Thus, solve \(\cos \theta = \frac{1}{4}\):

\(\theta = 75.5^\circ\) only.

(a) Start with the expression \(\left( \frac{1}{\cos x} - \tan x \right) \left( \frac{1}{\sin x} + 1 \right)\).

Use the identity \(\tan x = \frac{\sin x}{\cos x}\) to rewrite \(\tan x\) as \(\frac{\sin x}{\cos x}\).

Substitute to get \(\left( \frac{1}{\cos x} - \frac{\sin x}{\cos x} \right) \left( \frac{1}{\sin x} + 1 \right)\).

Simplify the first term: \(\frac{1 - \sin x}{\cos x}\).

Simplify the second term: \(\frac{1 + \sin x}{\sin x}\).

Multiply the two expressions: \(\frac{(1 - \sin x)(1 + \sin x)}{\cos x \sin x}\).

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

Thus, the expression becomes \(\frac{\cos^2 x}{\cos x \sin x} = \frac{\cos x}{\sin x}\).

Recognize \(\frac{\cos x}{\sin x} = \frac{1}{\tan x}\), proving the identity.

(b) From part (a), we have \(\frac{1}{\tan x} = 2 \tan^2 x\).

Rearrange to get \(\tan^3 x = \frac{1}{2}\).

Take the cube root: \(\tan x = \left( \frac{1}{2} \right)^{1/3}\).

Calculate \(x\) using \(\tan^{-1} \left( \left( \frac{1}{2} \right)^{1/3} \right)\).

Within the range \(0^\circ \leq x \leq 180^\circ\), \(x \approx 38.4^\circ\).

(a) Start by putting the expression over a single common denominator:

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

This simplifies to:

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

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

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

(b) From part (a), we have:

\(2 \tan^2 \theta = 8\)

\(\tan^2 \theta = 4\)

\(\tan \theta = \pm 2\)

Solving for \(\theta\) in the range \(0^\circ < \theta < 180^\circ\), we find:

\(\theta = 63.4^\circ, \ 116.6^\circ\)

(a) Start with the left-hand side:

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

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

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

\(= \frac{2}{\sin \theta \cos \theta}\)

Thus, the identity is proven.

(b) Given:

\(\frac{2}{\sin \theta \cos \theta} = \frac{6 \cos \theta}{\sin \theta}\)

\(\Rightarrow \frac{2}{\cos \theta} = 6\)

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

\(\Rightarrow \cos \theta = \pm 0.5774\)

For \(0^\circ < \theta < 180^\circ\), \(\theta = 54.7^\circ, 125.3^\circ\).