Past Exam Questions

To solve the problem, we follow these steps:

Part (i): Find \(\sin(a - 30^\circ)\)

We know \(\cos a = \frac{3}{5}\). Using the Pythagorean identity, \(\sin a = \sqrt{1 - \cos^2 a} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}\).

Using the angle subtraction formula for sine:

\(\sin(a - 30^\circ) = \sin a \cos 30^\circ - \cos a \sin 30^\circ\)

Substitute the known values:

\(\sin(a - 30^\circ) = \frac{4}{5} \cdot \frac{\sqrt{3}}{2} - \frac{3}{5} \cdot \frac{1}{2}\)

\(= \frac{4\sqrt{3}}{10} - \frac{3}{10}\)

\(= \frac{1}{10}(4\sqrt{3} - 3)\)

Part (ii): Find \(\tan 2a\) and \(\tan 3a\)

Using the double angle formula for tangent:

\(\tan 2a = \frac{2 \tan a}{1 - \tan^2 a}\)

First, find \(\tan a = \frac{\sin a}{\cos a} = \frac{4/5}{3/5} = \frac{4}{3}\).

Substitute into the formula:

\(\tan 2a = \frac{2 \cdot \frac{4}{3}}{1 - \left(\frac{4}{3}\right)^2} = \frac{\frac{8}{3}}{1 - \frac{16}{9}}\)

\(= \frac{\frac{8}{3}}{-\frac{7}{9}} = -\frac{24}{7}\)

Using the angle addition formula for tangent:

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

Substitute \(\tan 2a = -\frac{24}{7}\) and \(\tan a = \frac{4}{3}\):

\(\tan 3a = \frac{-\frac{24}{7} + \frac{4}{3}}{1 - \left(-\frac{24}{7}\right) \left(\frac{4}{3}\right)}\)

\(= \frac{-\frac{72}{21} + \frac{28}{21}}{1 + \frac{96}{21}}\)

\(= \frac{-\frac{44}{21}}{\frac{117}{21}} = -\frac{44}{117}\)

1. Given \(\tan \alpha = 2 \tan \beta\) and \(\tan(\alpha + \beta) = 3\).

2. Use the identity for \(\tan(A + B)\):

\(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = 3\)

3. Substitute \(\tan \alpha = 2 \tan \beta\) into the equation:

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

4. Simplify to get:

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

5. Solve for \(\tan \beta\):

\(3 \tan \beta = 3 - 6 \tan^2 \beta\)

\(6 \tan^2 \beta + 3 \tan \beta - 3 = 0\)

6. Divide by 3:

\(2 \tan^2 \beta + \tan \beta - 1 = 0\)

7. Solve the quadratic equation:

\(\tan \beta = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\)

8. Solutions are \(\tan \beta = \frac{1}{2}\) or \(\tan \beta = -1\).

9. For \(\tan \beta = \frac{1}{2}\), \(\beta = 26.6^\circ\) and \(\tan \alpha = 1\), \(\alpha = 45^\circ\).

10. For \(\tan \beta = -1\), \(\beta = 135^\circ\) and \(\tan \alpha = -2\), \(\alpha = 116.6^\circ\).

(i) Use the tangent addition and subtraction formulas:

\(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)

\(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\)

Substitute \(A = 30^\circ\), \(B = \theta\), and \(A = 60^\circ\), \(B = \theta\):

\(\tan(30^\circ + \theta) = \frac{\tan 30^\circ + \tan \theta}{1 - \tan 30^\circ \tan \theta}\)

\(\tan(60^\circ - \theta) = \frac{\tan 60^\circ - \tan \theta}{1 + \tan 60^\circ \tan \theta}\)

Given \(\tan 30^\circ = \frac{1}{\sqrt{3}}\) and \(\tan 60^\circ = \sqrt{3}\), substitute these values:

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

Cross-multiply and simplify:

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

After simplification, you obtain:

\(\tan^2 \theta + 6\sqrt{3} \tan \theta - 5 = 0\)

(ii) Solve the quadratic equation:

\(\tan^2 \theta + 6\sqrt{3} \tan \theta - 5 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 6\sqrt{3}\), \(c = -5\):

\(\tan \theta = \frac{-6\sqrt{3} \pm \sqrt{(6\sqrt{3})^2 - 4 \times 1 \times (-5)}}{2 \times 1}\)

\(\tan \theta = \frac{-6\sqrt{3} \pm \sqrt{108 + 20}}{2}\)

\(\tan \theta = \frac{-6\sqrt{3} \pm \sqrt{128}}{2}\)

\(\tan \theta = \frac{-6\sqrt{3} \pm 8\sqrt{2}}{2}\)

Calculate \(\theta\) for \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = 24.7^\circ\) and \(\theta = 95.3^\circ\).

(i) Use the tangent addition formula: \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\).

For \(\tan(45^\circ + x)\):

\(\tan(45^\circ + x) = \frac{\tan 45^\circ + \tan x}{1 - \tan 45^\circ \tan x} = \frac{1 + \tan x}{1 - \tan x}\)

Substitute into the equation:

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

Multiply through by \(1 - \tan x\):

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

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

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

Rearrange to obtain:

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

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

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

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

\(\tan x = \frac{-2 \pm \sqrt{4 + 4}}{2}\)

\(\tan x = \frac{-2 \pm \sqrt{8}}{2}\)

\(\tan x = \frac{-2 \pm 2\sqrt{2}}{2}\)

\(\tan x = -1 \pm \sqrt{2}\)

For \(\tan x = -1 + \sqrt{2}\), \(x = 22.5^\circ\).

For \(\tan x = -1 - \sqrt{2}\), \(x = 112.5^\circ\).

Thus, the solutions are \(x = 22.5^\circ\) and \(x = 112.5^\circ\).

(i) Use the tangent addition and subtraction formulas:

\(\tan(45^\circ + x) = \frac{1 + \tan x}{1 - \tan x}\)

\(\tan(45^\circ - x) = \frac{1 - \tan x}{1 + \tan x}\)

Substitute into the equation:

\(\frac{1 + \tan x}{1 - \tan x} = 2 \cdot \frac{1 - \tan x}{1 + \tan x}\)

Cross-multiply to clear the fractions:

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

Expand both sides:

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

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

Rearrange to form a quadratic equation:

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

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

Divide by 3:

\(\tan^2 x - 6\tan x + 1 = 0\)

(ii) Solve the quadratic equation \(\tan^2 x - 6\tan x + 1 = 0\):

Let \(y = \tan x\), then:

\(y^2 - 6y + 1 = 0\)

Use the quadratic formula:

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

\(y = \frac{6 \pm \sqrt{36 - 4}}{2}\)

\(y = \frac{6 \pm \sqrt{32}}{2}\)

\(y = \frac{6 \pm 4\sqrt{2}}{2}\)

\(y = 3 \pm 2\sqrt{2}\)

Calculate \(x\) for \(y = \tan x\):

\(x = \tan^{-1}(3 + 2\sqrt{2}) \approx 80.3^\circ\)

\(x = \tan^{-1}(3 - 2\sqrt{2}) \approx 9.7^\circ\)

(i) Start with the equation:

\(\sin(x - 60^\circ) - \cos(30^\circ - x) = 1\)

Use the identity \(\sin(a - b) = \sin a \cos b - \cos a \sin b\):

\(\sin(x - 60^\circ) = \sin x \cos 60^\circ - \cos x \sin 60^\circ\)

\(\cos(30^\circ - x) = \cos 30^\circ \cos x + \sin 30^\circ \sin x\)

Substitute these into the equation:

\((\sin x \cos 60^\circ - \cos x \sin 60^\circ) - (\cos 30^\circ \cos x + \sin 30^\circ \sin x) = 1\)

Using \(\cos 60^\circ = \frac{1}{2}\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\sin 30^\circ = \frac{1}{2}\):

\(\left(\sin x \cdot \frac{1}{2} - \cos x \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) = 1\)

Simplify:

\(\sin x \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cos x - \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = 1\)

\(-\sqrt{3} \cos x = 1\)

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

Thus, \(k = -\frac{1}{\sqrt{3}}\).

(ii) Solve \(\cos x = -\frac{1}{\sqrt{3}}\) for \(0^\circ < x < 180^\circ\):

The angle \(x\) where \(\cos x = -\frac{1}{\sqrt{3}}\) is approximately \(x = 125.3^\circ\).

Given:

\(\tan(\alpha + \beta) = 2\)

\(\tan \alpha = 3 \tan \beta\)

Using the tangent addition formula:

\(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = 2\)

Substitute \(\tan \alpha = 3 \tan \beta\):

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

Simplify:

\(\frac{4 \tan \beta}{1 - 3 \tan^2 \beta} = 2\)

Cross-multiply:

\(4 \tan \beta = 2(1 - 3 \tan^2 \beta)\)

\(4 \tan \beta = 2 - 6 \tan^2 \beta\)

Rearrange to form a quadratic equation:

\(6 \tan^2 \beta + 4 \tan \beta - 2 = 0\)

Divide by 2:

\(3 \tan^2 \beta + 2 \tan \beta - 1 = 0\)

Let \(x = \tan \beta\), solve:

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

Using the quadratic formula:

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

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

\(x = \frac{-2 \pm \sqrt{4 + 12}}{6}\)

\(x = \frac{-2 \pm \sqrt{16}}{6}\)

\(x = \frac{-2 \pm 4}{6}\)

\(x = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = \frac{-6}{6} = -1\)

Thus, \(\tan \beta = \frac{1}{3}\) or \(\tan \beta = -1\).

For \(\tan \beta = \frac{1}{3}\), \(\tan \alpha = 1\), so \(\alpha = 45^\circ\) and \(\beta = 18.4^\circ\).

For \(\tan \beta = -1\), \(\tan \alpha = -3\), so \(\alpha = 108.4^\circ\) and \(\beta = 135^\circ\).

(a) Start by using the double angle formula:

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

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

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

Simplify to get:

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

Set \(\tan 4\theta = \frac{1}{2} \tan \theta\) and simplify:

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

Cancel \(\tan \theta\) and solve:

\(8 (1 - \tan^2 \theta) = (1 - \tan^2 \theta)^2 - 4 \tan^2 \theta\)

Rearrange and simplify to get:

\(\tan^4 \theta + 2 \tan^2 \theta - 7 = 0\)

(b) Solve \(\tan^4 \theta + 2 \tan^2 \theta - 7 = 0\) by substituting \(x = \tan^2 \theta\):

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

Use the quadratic formula:

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-7)}}{2 \times 1}\)

\(x = \frac{-2 \pm \sqrt{4 + 28}}{2}\)

\(x = \frac{-2 \pm \sqrt{32}}{2}\)

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

\(x = -1 \pm 2\sqrt{2}\)

Only positive \(x\) is valid: \(x = -1 + 2\sqrt{2}\)

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

\(\tan \theta = \sqrt{-1 + 2\sqrt{2}}\)

Calculate \(\theta\) for \(0^\circ < \theta < 180^\circ\):

\(\theta = 53.5^\circ, 126.5^\circ\)

(a) Start with the given equation:

\(\cos(x - 30^\circ) = 2 \sin(x + 30^\circ)\)

Use the angle subtraction and addition formulas:

\(\cos(x - 30^\circ) = \cos x \cos 30^\circ + \sin x \sin 30^\circ\)

\(\sin(x + 30^\circ) = \sin x \cos 30^\circ + \cos x \sin 30^\circ\)

Substitute the exact values:

\(\cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}\)

\(\cos(x - 30^\circ) = \cos x \frac{\sqrt{3}}{2} + \sin x \frac{1}{2}\)

\(\sin(x + 30^\circ) = \sin x \frac{\sqrt{3}}{2} + \cos x \frac{1}{2}\)

Substitute into the original equation:

\(\cos x \frac{\sqrt{3}}{2} + \sin x \frac{1}{2} = 2 \left( \sin x \frac{\sqrt{3}}{2} + \cos x \frac{1}{2} \right)\)

Expand and simplify:

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

Rearrange terms:

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

Factor out terms:

\(\cos x \left( \frac{\sqrt{3}}{2} - 1 \right) = \sin x \left( \sqrt{3} - \frac{1}{2} \right)\)

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

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

Simplify to obtain:

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

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

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

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

\(x = \tan^{-1}\left(\frac{2 - \sqrt{3}}{1 - 2\sqrt{3}}\right)\)

Find solutions within the interval \(0^\circ < x < 360^\circ\):

\(x \approx 173.8^\circ, 353.8^\circ\)

1. Use the identity \(\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}\) to express \(\tan(x + 45^\circ)\):

\(\tan(x + 45^\circ) = \frac{\tan x + 1}{1 - \tan x}\)

2. Substitute \(\cot x = \frac{1}{\tan x}\) into the equation:

\(\frac{\tan x + 1}{1 - \tan x} = 2 \cdot \frac{1}{\tan x} + 1\)

3. Clear the fractions by multiplying through by \(\tan x (1 - \tan x)\):

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

4. Simplify and rearrange to form a quadratic equation:

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

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

5. Solve the quadratic equation using the quadratic formula \(\tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

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

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

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

\(\tan x = \frac{-1 \pm 3}{4}\)

\(\tan x = \frac{2}{4} = 0.5 \quad \text{or} \quad \tan x = \frac{-4}{4} = -1\)

6. Find \(x\) for \(\tan x = 0.5\):

\(x = \tan^{-1}(0.5) \approx 26.6^\circ\)

7. Find \(x\) for \(\tan x = -1\):

\(x = 135^\circ\)

8. Adjust for the given interval \(0^\circ < x < 180^\circ\):

\(x = 31.7^\circ \quad \text{and} \quad x = 121.7^\circ\)

(a) Start with the identity for tangent of a sum:

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \tan 60^\circ}{1 - \tan \theta \tan 60^\circ}\)

Since \(\tan 60^\circ = \sqrt{3}\), substitute to get:

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}}\)

Given \(\tan(\theta + 60^\circ) = 2 \cot \theta\), rewrite \(\cot \theta\) as \(\frac{1}{\tan \theta}\):

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

Cross-multiply to obtain:

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

Expand and simplify:

\(\tan^2 \theta + \sqrt{3} \tan \theta = 2 - 2\sqrt{3} \tan \theta\)

Rearrange to form the quadratic equation:

\(\tan^2 \theta + 3\sqrt{3} \tan \theta - 2 = 0\)

(b) Solve the quadratic equation \(\tan^2 \theta + 3\sqrt{3} \tan \theta - 2 = 0\) using the quadratic formula:

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

Calculate the discriminant:

\(\sqrt{27 + 8} = \sqrt{35}\)

Thus,

\(\tan \theta = \frac{-3\sqrt{3} \pm \sqrt{35}}{2}\)

Calculate the solutions for \(\theta\):

\(\tan \theta = 0.3599\) gives \(\theta = 19.8^\circ\)

\(\tan \theta = -5.556\) gives \(\theta = 100.2^\circ\)

Thus, the solutions are \(\theta = 19.8^\circ\) and \(\theta = 100.2^\circ\).

(a) Start with the given equation:

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

Use the identity \(\cot^2 \theta = \csc^2 \theta - 1\):

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

Substitute \(\csc^2 \theta = \frac{1}{\sin^2 \theta}\) and \(\cos 2\theta = 1 - 2 \sin^2 \theta\):

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

Simplify:

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

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

Multiply through by \(\sin^2 \theta\):

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

Rearrange to:

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

(b) Solve the quadratic equation:

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

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

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

\(x = \frac{-3 \pm \sqrt{3^2 - 4 \times 4 \times (-1)}}{2 \times 4}\)

\(x = \frac{-3 \pm \sqrt{9 + 16}}{8}\)

\(x = \frac{-3 \pm 5}{8}\)

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

Since \(x = \sin^2 \theta\), \(x = -1\) is not valid. So, \(\sin^2 \theta = \frac{1}{4}\).

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

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

To solve \(\cot 2\theta = 2 \tan \theta\), we start by using trigonometric identities:

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

Equating this to \(2 \tan \theta\):

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

Cross-multiply to clear the fraction:

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

Rearrange the equation:

\(1 = 5 \tan^2 \theta\)

\(\tan^2 \theta = \frac{1}{5}\)

Take the square root:

\(\tan \theta = \pm \frac{1}{\sqrt{5}}\)

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

\(\theta = \tan^{-1} \left( \frac{1}{\sqrt{5}} \right) \approx 24.1^\circ\)

For the second solution in the range \(0^\circ < \theta < 180^\circ\):

\(\theta = 180^\circ - 24.1^\circ = 155.9^\circ\)

Thus, the solutions are \(\theta = 24.1^\circ\) and \(\theta = 155.9^\circ\).

(i) Start with the identity \(\cos^2 x + \sin^2 x = 1\). Raise both sides to the power of 3:

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

Expand \((\cos^2 x + \sin^2 x)^3\):

\((\cos^2 x + \sin^2 x)^3 = \cos^6 x + 3\cos^4 x \sin^2 x + 3\cos^2 x \sin^4 x + \sin^6 x\).

Using \(\cos^2 x + \sin^2 x = 1\), substitute \(\cos^2 x = 1 - \sin^2 x\) and \(\sin^2 x = 1 - \cos^2 x\):

\(\cos^6 x + \sin^6 x = 1 - 3\cos^2 x \sin^2 x\).

Using the double angle identity \(\sin 2x = 2 \sin x \cos x\), we have \(\sin^2 2x = 4 \sin^2 x \cos^2 x\):

\(\cos^6 x + \sin^6 x = 1 - \frac{3}{4} \sin^2 2x\).

(ii) Given \(\cos^6 x + \sin^6 x = \frac{2}{3}\), substitute the result from part (i):

\(1 - \frac{3}{4} \sin^2 2x = \frac{2}{3}\).

Solve for \(\sin^2 2x\):

\(\frac{3}{4} \sin^2 2x = 1 - \frac{2}{3} = \frac{1}{3}\).

\(\sin^2 2x = \frac{4}{9}\).

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

For \(\sin 2x = \frac{2}{3}\), solve for \(2x\):

\(2x = \sin^{-1}\left(\frac{2}{3}\right)\) gives \(2x \approx 41.8^\circ\) or \(2x \approx 138.2^\circ\).

Thus, \(x \approx 20.9^\circ\) or \(x \approx 69.1^\circ\).

For \(\sin 2x = -\frac{2}{3}\), solve for \(2x\):

\(2x = \sin^{-1}\left(-\frac{2}{3}\right)\) gives \(2x \approx 221.8^\circ\) or \(2x \approx 318.2^\circ\).

Thus, \(x \approx 110.9^\circ\) or \(x \approx 159.1^\circ\).

Therefore, the solutions are \(x = 20.9^\circ, 69.1^\circ, 110.9^\circ, 159.1^\circ\).

Step 1: Express \(\cot \theta\) and \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).

\(\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}\)

Step 2: Substitute into the equation:

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

Step 3: Use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\):

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

Step 4: Multiply through by \(\sin \theta \cos \theta\) to clear the fraction:

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

Step 5: Use \(\sin^2 \theta = 1 - \cos^2 \theta\):

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

Step 6: Simplify:

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

Step 7: Rearrange to form a quadratic in \(\cos^2 \theta\):

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

Step 8: Solve for \(\cos^2 \theta\):

Let \(x = \cos^2 \theta\), then:

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

Step 9: Solve the quadratic equation using the quadratic formula:

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

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

\(x = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4}\)

Step 10: Find \(\theta\) for \(90^\circ < \theta < 180^\circ\):

Only the positive root is valid for \(\cos^2 \theta\), so:

\(\cos^2 \theta = \frac{-1 + \sqrt{17}}{4}\)

\(\theta = 152.1^\circ\)

To express \(\cot 2\theta\) in terms of \(\tan \theta\), use the identity:

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

Given the equation:

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

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

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

Rearrange to form a quadratic equation:

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

Let \(x = \tan \theta\). The equation becomes:

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

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

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

\(x = \frac{-2 \pm \sqrt{4 + 12}}{6}\)

\(x = \frac{-2 \pm \sqrt{16}}{6}\)

\(x = \frac{-2 \pm 4}{6}\)

\(x = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = \frac{-6}{6} = -1\)

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

For \(\tan \theta = \frac{1}{3}\), \(\theta = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.4^\circ\).

For \(\tan \theta = -1\), \(\theta = 135^\circ\) (since \(\tan 135^\circ = -1\)).

Thus, the solutions are \(\theta = 18.4^\circ\) and \(\theta = 135^\circ\).

To express \(\sec \theta = 3 \cos \theta + \tan \theta\) as a quadratic in \(\sin \theta\), we start by using trigonometric identities:

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

Substitute these into the equation:

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

Multiply through by \(\cos \theta\) to clear the fractions:

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

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

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

Expand and rearrange:

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

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

This is a quadratic equation in \(\sin \theta\). Let \(x = \sin \theta\), then:

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

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

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

\(x = \frac{1 \pm \sqrt{1 + 24}}{6}\)

\(x = \frac{1 \pm 5}{6}\)

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

Since \(x = \sin \theta\), and \(-1 \leq \sin \theta \leq 1\), only \(x = -\frac{2}{3}\) is valid in the interval \(-90^\circ < \theta < 90^\circ\).

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

Find \(\theta\):

\(\theta = \arcsin\left(-\frac{2}{3}\right) \approx -41.8^\circ\)

(i) To prove the identity \(\cos 4\theta - 4\cos 2\theta \equiv 8\sin^4\theta - 3\):

Use double angle formulas:

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

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

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

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

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

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

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

\(\cos 4\theta - 4\cos 2\theta = 8\sin^4 \theta - 8\sin^2 \theta + 1 - 4(1 - 2\sin^2 \theta)\)

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

\(= 8\sin^4 \theta - 3\)

Thus, the identity is proven.

(ii) To solve \(\cos 4\theta = 4\cos 2\theta + 3\):

Using the identity \(\cos 4\theta - 4\cos 2\theta = 8\sin^4 \theta - 3\), set \(8\sin^4 \theta - 3 = 3\).

\(8\sin^4 \theta = 6\)

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

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

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

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

Adjust for the equation \(\cos 4\theta = 4\cos 2\theta + 3\):

\(\theta = 68.5^\circ, 111.5^\circ, 248.5^\circ, 291.5^\circ\)

1. Start with the given equation:

\(\csc \theta = 3 \sin \theta + \cot \theta\)

2. Use the identities \(\csc \theta = \frac{1}{\sin \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\):

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

3. Multiply through by \(\sin \theta\) to clear the fractions:

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

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

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

5. Simplify and rearrange to form a quadratic equation:

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

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

6. Solve the quadratic equation \(3\cos^2 \theta - \cos \theta - 2 = 0\) using the quadratic formula \(\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -1\), \(c = -2\):

\(\cos \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 3 \times (-2)}}{2 \times 3}\)

\(\cos \theta = \frac{1 \pm \sqrt{1 + 24}}{6}\)

\(\cos \theta = \frac{1 \pm 5}{6}\)

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

7. Since \(\cos \theta = 1\) is not in the interval \(0^\circ < \theta < 180^\circ\), we use \(\cos \theta = -\frac{2}{3}\):

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

1. Use the identity \(\cot 2x = \frac{\cot^2 x - 1}{2\cot x}\) to express \(\cot 2x\) in terms of \(\cot x\).

2. Substitute into the equation: \(\frac{\cot^2 x - 1}{2\cot x} + \cot x = 3\).

3. Multiply through by \(2\cot x\) to clear the fraction: \(\cot^2 x - 1 + 2\cot^2 x = 6\cot x\).

4. Simplify to get: \(3\cot^2 x - 6\cot x - 1 = 0\).

5. Let \(y = \cot x\), then the equation becomes \(3y^2 - 6y - 1 = 0\).

6. Solve the quadratic equation using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3, b = -6, c = -1\).

7. Calculate the discriminant: \(b^2 - 4ac = 36 + 12 = 48\).

8. Solve for \(y\): \(y = \frac{6 \pm \sqrt{48}}{6}\).

9. Simplify to find \(y = 3 \pm \sqrt{3}\).

10. Find \(x\) using \(x = \cot^{-1}(y)\).

11. Calculate \(x\) for \(y = 3 + \sqrt{3}\) and \(y = 3 - \sqrt{3}\) within the interval \(0^\circ < x < 180^\circ\).

12. The solutions are \(x = 24.9^\circ\) and \(x = 98.8^\circ\).