Past Exam Questions

(i) Start by expanding \(\tan(2x + x)\) using the angle addition formula:

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

Using \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), substitute into the equation:

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

Simplify the expression:

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

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

Given \(\tan 3x = 3 \cot x = \frac{3}{\tan x}\), equate and simplify:

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

Cross-multiply and simplify:

\(3 \tan^2 x - \tan^4 x = 3 - 6 \tan^2 x\)

Rearrange to obtain:

\(\tan^4 x - 12 \tan^2 x + 3 = 0\)

(ii) Solve \(\tan^4 x - 12 \tan^2 x + 3 = 0\) by substituting \(y = \tan^2 x\):

\(y^2 - 12y + 3 = 0\)

Use the quadratic formula:

\(y = \frac{12 \pm \sqrt{144 - 12}}{2}\)

\(y = \frac{12 \pm \sqrt{132}}{2}\)

\(y = 6 \pm \sqrt{33}\)

Since \(y = \tan^2 x\), solve for \(x\):

\(\tan x = \sqrt{6 + \sqrt{33}}\) or \(\tan x = \sqrt{6 - \sqrt{33}}\)

Calculate \(x\) for \(0^\circ < x < 90^\circ\):

\(x = 26.8^\circ\) and \(x = 73.7^\circ\)

1. Start with the equation:

\(\cot \theta - \cot(\theta + 45^\circ) = 3\)

2. Use the identity \(\cot(\theta + 45^\circ) = \frac{1 - \tan \theta}{\tan \theta + 1}\).

3. Substitute \(\cot \theta = \frac{1}{\tan \theta}\) and \(\cot(\theta + 45^\circ) = \frac{1 - \tan \theta}{\tan \theta + 1}\) into the equation:

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

4. Simplify and rearrange to form a quadratic equation:

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

5. Expand and simplify:

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

6. Solve the quadratic equation using the quadratic formula:

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

where \(a = 2\), \(b = 3\), \(c = -1\).

7. Calculate:

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

\(\tan \theta = \frac{-3 \pm \sqrt{9 + 8}}{4}\)

\(\tan \theta = \frac{-3 \pm \sqrt{17}}{4}\)

8. Find \(\theta\) for \(0^\circ < \theta < 180^\circ\):

\(\theta = 15.7^\circ\) or \(\theta = 119.3^\circ\).

(i) Start with the given equation:

\(\sin(\theta + 45^\circ) + 2 \cos(\theta + 60^\circ) = 3 \cos \theta\)

Use angle addition formulas:

\(\sin(\theta + 45^\circ) = \sin \theta \cos 45^\circ + \cos \theta \sin 45^\circ\)

\(\cos(\theta + 60^\circ) = \cos \theta \cos 60^\circ - \sin \theta \sin 60^\circ\)

Substitute \(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\), \(\cos 60^\circ = \frac{1}{2}\), and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):

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

Simplify:

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

Combine terms:

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

Rearrange:

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

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

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

\(\tan \theta = \frac{2\sqrt{2} - 1}{1 - \sqrt{6}}\)

(ii) Solve for \(\theta\):

Using the expression for \(\tan \theta\), find \(\theta\) in the range \(0^\circ < \theta < 360^\circ\).

\(\theta = 128.4^\circ\) and \(\theta = 308.4^\circ\)

To solve the equation \(\sin(\theta - 30^\circ) + \cos \theta = 2 \sin \theta\), we start by using the angle subtraction identity:

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

Substitute the values \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\):

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

Simplify the equation:

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

Combine like terms:

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

Rearrange to form:

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

Factor out \(\sin \theta\):

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

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

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

Calculate \(\theta\):

\(\theta = \tan^{-1}\left(\frac{1}{4 - \sqrt{3}}\right) \approx 23.8^\circ\)

Thus, the solution is \(\theta = 23.8^\circ\).

We solve the equation \(\cot \theta + \cot(\theta + 45^\circ) = 2\) using the addition formula for tangent.

Let \(t = \tan \theta\). Using the identity

\[ \tan(\theta + 45^\circ) = \frac{\tan \theta + \tan 45^\circ}{1 - \tan \theta \tan 45^\circ} = \frac{t + 1}{1 - t}, \] so \[ \cot(\theta + 45^\circ) = \frac{1}{\tan(\theta + 45^\circ)} = \frac{1 - t}{t + 1}. \]

Also, \(\cot \theta = \frac{1}{t}\). Substituting these into the original equation gives

\[ \frac{1}{t} + \frac{1 - t}{t + 1} = 2. \]

Multiply through by \(t(t+1)\):

\[ (t+1) + t(1 - t) = 2t(t+1). \]

Expanding and simplifying:

\[ t + 1 + t - t^2 = 2t^2 + 2t \quad \Rightarrow \quad 2t - t^2 + 1 = 2t^2 + 2t. \]

Bring all terms to one side:

\[ 0 = 3t^2 - 1 \quad \Rightarrow \quad t^2 = \frac{1}{3}. \]

So

\[ t = \pm \frac{1}{\sqrt{3}}. \]

This gives \(\tan \theta = \frac{1}{\sqrt{3}}\) or \(\tan \theta = -\frac{1}{\sqrt{3}}\).

In degrees:

• \(\tan \theta = \frac{1}{\sqrt{3}}\) gives \(\theta = 30^\circ + 180^\circ k\).
• \(\tan \theta = -\frac{1}{\sqrt{3}}\) gives \(\theta = 150^\circ + 180^\circ k\).

For \(0^\circ < \theta < 180^\circ\), the solutions are \(\boxed{\theta = 30^\circ \text{ and } \theta = 150^\circ}\).

Part (i):

Given the equation:

\(\sin(x - 60^\circ) = 3 \cos(x - 45^\circ)\)

Using the angle subtraction identities:

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

\(\cos(x - 45^\circ) = \cos x \cos 45^\circ + \sin x \sin 45^\circ\)

Substitute the exact values:

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

Rearrange and simplify:

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

Combine terms:

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

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

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

\(\tan x = \frac{-(6+\sqrt{6})}{(6-\sqrt{2})}\)

Part (ii):

Using the result from part (i), solve for \(x\):

\(x = 118.5^\circ, 298.5^\circ\)

(i) To prove the identity \(\tan(45^\circ + x) + \tan(45^\circ - x) \equiv 2 \sec 2x\):

Use the tangent addition and subtraction formulas:

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

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

Add the two expressions:

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

Combine into a single fraction:

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

Simplify the numerator:

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

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

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

Thus, the identity is proven.

(ii) To sketch the graph of \(y = \tan(45^\circ + x) + \tan(45^\circ - x)\) for \(0^\circ \leq x \leq 90^\circ\):

The function has an asymptote at \(x = 45^\circ\) because \(\tan(45^\circ + x)\) and \(\tan(45^\circ - x)\) become undefined.

The graph consists of two branches approaching the asymptote at \(x = 45^\circ\).

1. Use the angle addition and subtraction formulas:

\(\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)

2. Apply the formulas:

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

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

3. Substitute into the equation:

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

4. Simplify and solve for \(\tan \theta\):

Multiply through by \((1 - \tan \theta)(1 + \tan \theta)\):

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

5. Simplify further:

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

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

6. Rearrange to form a quadratic equation:

\(7\tan^2 \theta - 2\tan \theta - 1 = 0\)

7. Solve the quadratic equation using the quadratic formula:

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

where \(a = 7, b = -2, c = -1\).

8. Calculate:

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

\(\tan \theta = \frac{2 \pm \sqrt{4 + 28}}{14}\)

\(\tan \theta = \frac{2 \pm \sqrt{32}}{14}\)

\(\tan \theta = \frac{2 \pm 4\sqrt{2}}{14}\)

\(\tan \theta = \frac{1 \pm 2\sqrt{2}}{7}\)

9. Find \(\theta\) using \(\tan^{-1}\):

\(\theta = 28.7^\circ\) and \(\theta = 165.4^\circ\).

1. Use the identity for \(\sin(A + 45^\circ)\):

\(\sin A \cos 45^\circ + \cos A \sin 45^\circ = 2\sqrt{2} \cos A\)

2. Simplify using \(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\):

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

3. Divide through by \(\cos A\):

\(\tan A \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 2\sqrt{2}\)

4. Solve for \(\tan A\):

\(\tan A = 3\)

5. Use the identity \(\sec^2 B = 1 + \tan^2 B\) in the second equation:

\(4(1 + \tan^2 B) + 5 = 12 \tan B\)

6. Simplify and solve the quadratic equation:

\(4\tan^2 B - 12\tan B + 9 = 0\)

7. Factor or use the quadratic formula to find:

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

8. Use the identity for \(\tan(A - B)\):

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

9. Substitute \(\tan A = 3\) and \(\tan B = \frac{3}{2}\):

\(\tan(A - B) = \frac{3 - \frac{3}{2}}{1 + 3 \cdot \frac{3}{2}}\)

10. Simplify:

\(\tan(A - B) = \frac{\frac{3}{2}}{1 + \frac{9}{2}} = \frac{\frac{3}{2}}{\frac{11}{2}} = \frac{3}{11}\)

Given:

\(\tan(\theta - \phi) = 3\)

\(\tan \theta + \tan \phi = 1\)

Using the tangent subtraction formula:

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

Substitute \(\tan \theta + \tan \phi = 1\) into the equation:

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

Simplify:

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

Cross-multiply and simplify:

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

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

Rearrange to form a quadratic equation:

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

Solving this quadratic equation gives:

\(\tan \theta = 1.5 \quad \text{or} \quad \tan \theta = -\frac{4}{3}\)

For \(\tan \theta = 1.5\), \(\theta \approx 56.3^\circ\) or \(236.3^\circ\) (only \(56.3^\circ\) is valid).

For \(\tan \theta = -\frac{4}{3}\), \(\theta \approx 126.9^\circ\) or \(306.9^\circ\) (only \(126.9^\circ\) is valid).

Using \(\tan \theta + \tan \phi = 1\), find \(\phi\):

For \(\theta = 56.3^\circ\), \(\phi \approx 161.6^\circ\).

For \(\theta = 126.9^\circ\), \(\phi \approx 63.4^\circ\).

Thus, the possible pairs are \((\theta, \phi) = (135^\circ, 63.4^\circ)\) and \((53.1^\circ, 161.6^\circ)\).

(i) Use the sum-to-product identities:

\(\cos(\theta - 60^\circ) + \cos(\theta + 60^\circ) = 2 \cos \left( \frac{2\theta}{2} \right) \cos \left( \frac{-120^\circ}{2} \right)\)

\(= 2 \cos \theta \cos(-60^\circ)\)

Since \(\cos(-60^\circ) = \cos(60^\circ) = \frac{1}{2}\), we have:

\(2 \cos \theta \cdot \frac{1}{2} = \cos \theta\)

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

(ii) Start with:

\(\frac{\cos(2x - 60^\circ) + \cos(2x + 60^\circ)}{\cos(x - 60^\circ) + \cos(x + 60^\circ)} = 3\)

Using the identity \(\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)\), we have:

\(\cos(2x - 60^\circ) + \cos(2x + 60^\circ) = 2 \cos(2x) \cos(60^\circ) = \cos(2x)\)

\(\cos(x - 60^\circ) + \cos(x + 60^\circ) = 2 \cos(x) \cos(60^\circ) = \cos(x)\)

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

Using the double angle identity \(\cos(2x) = 2 \cos^2(x) - 1\), we get:

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

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

Solving this quadratic equation for \(\cos(x)\):

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

\(2y^2 - 3y - 1 = 0\)

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

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

\(y = \frac{3 \pm \sqrt{9 + 8}}{4}\)

\(y = \frac{3 \pm \sqrt{17}}{4}\)

Since \(\cos(x)\) must be between -1 and 1, the valid solution is:

\(\cos(x) = \frac{1}{4}(3 - \sqrt{17})\)

(i) Expand \(\sin(2\theta + \theta)\) using the angle addition formula:

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

Using double angle identities:

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

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

Substitute these into the expansion:

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

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

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

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

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

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

This proves the identity.

(ii) Substitute \(x = \frac{2 \sin \theta}{\sqrt{3}}\) into the equation:

\(x^3 - x + \frac{1}{6}\sqrt{3} = 0\)

\(\left(\frac{2 \sin \theta}{\sqrt{3}}\right)^3 - \frac{2 \sin \theta}{\sqrt{3}} + \frac{1}{6}\sqrt{3} = 0\)

\(\frac{8 \sin^3 \theta}{3\sqrt{3}} - \frac{2 \sin \theta}{\sqrt{3}} + \frac{1}{6}\sqrt{3} = 0\)

Multiply through by \(3\sqrt{3}\):

\(8 \sin^3 \theta - 6 \sin \theta + \frac{1}{2} \cdot 3 = 0\)

\(8 \sin^3 \theta - 6 \sin \theta + \frac{3}{2} = 0\)

\(8 \sin^3 \theta - 6 \sin \theta = -\frac{3}{2}\)

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

This confirms the form \(\sin 3\theta = \frac{3}{4}\).

(iii) Solve \(x^3 - x + \frac{1}{6}\sqrt{3} = 0\):

Using numerical methods or a calculator, find:

\(x = 0.322, 0.799, -1.12\)

Part (i):

Start with the given equation:

\(\tan(x - 60^\circ) + \cot x = \sqrt{3}\)

Using the identity \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\), we have:

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

Substitute into the equation:

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

Multiply through by \(\tan x (1 + \sqrt{3} \tan x)\):

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

Simplify and rearrange:

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

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

Rearrange to form:

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

Part (ii):

Solve the quadratic equation:

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

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

\(2u^2 + \sqrt{3}u - 1 = 0\)

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

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

\(u = \frac{-\sqrt{3} \pm \sqrt{3 + 8}}{4}\)

\(u = \frac{-\sqrt{3} \pm \sqrt{11}}{4}\)

Calculate \(\tan x\) and find \(x\):

\(x = 21.6^\circ\) and \(x = 128.4^\circ\)

To solve the equation \(\cos(x + 30^\circ) = 2 \cos x\), we use the cosine addition formula:

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

Substitute \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\):

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

Rearrange to form:

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

Simplify:

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

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

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

Simplify the expression:

\(\tan x = \sqrt{3} - 4.\)

Find \(x\) using the arctangent function:

\(x = \tan^{-1}(\sqrt{3} - 4).\)

Calculate the solutions within the interval \(-180^\circ < x < 180^\circ\):

\(x \approx -66.2^\circ \text{ and } x \approx 113.8^\circ.\)

To solve \(\sin(\theta + 45^\circ) = 2 \cos(\theta - 30^\circ)\), we use the angle addition formulas:

\(\sin(\theta + 45^\circ) = \sin \theta \cos 45^\circ + \cos \theta \sin 45^\circ\)

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

Substitute these into the equation:

\(\sin \theta \cos 45^\circ + \cos \theta \sin 45^\circ = 2(\cos \theta \cos 30^\circ + \sin \theta \sin 30^\circ)\)

Using \(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), and \(\sin 30^\circ = \frac{1}{2}\), we have:

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

Simplify:

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

Rearrange terms:

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

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

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

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

Calculate \(\theta\):

\(\theta = \tan^{-1}\left( \frac{\sqrt{3} - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} - 1} \right) \approx 105.9^\circ\)

Thus, the solution is \(\theta = 105.9^\circ\).

(i) Start with \(\tan(3x) = \tan(2x + x)\).

Using the identity \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\), we have:

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

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

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

Simplify the expression:

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

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

Equating to \(k \tan x\), we get:

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

\(3 \tan x - \tan^3 x = k \tan x (1 - 2 \tan^2 x)\)

\(3 \tan x - \tan^3 x = k \tan x - 2k \tan^3 x\)

Rearranging gives:

\((3k - 1) \tan^2 x = k - 3\)

(ii) Substitute \(k = 4\) into \((3k - 1) \tan^2 x = k - 3\):

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

\(11 \tan^2 x = 1\)

\(\tan^2 x = \frac{1}{11}\)

\(\tan x = \pm \frac{1}{\sqrt{11}}\)

Find \(x\) using \(\tan^{-1}\):

\(x = \tan^{-1}\left(\frac{1}{\sqrt{11}}\right) \approx 16.8^\circ\)

\(x = 180^\circ - 16.8^\circ = 163.2^\circ\)

(iii) For \(k = 2\), substitute into \((3k - 1) \tan^2 x = k - 3\):

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

\(5 \tan^2 x = -1\)

\(\tan^2 x\) cannot be negative, so no solutions exist.

(i) Use the tangent addition and subtraction formulas:

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

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

Given \(\tan 60^\circ = \sqrt{3}\), substitute:

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

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

Add the two equations:

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

Simplify the numerator:

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

Simplify the denominator:

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

Thus, the equation becomes:

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

Therefore, \((2\sqrt{3})(1 + \tan^2 \theta) = k(1 - 3\tan^2 \theta)\).

(ii) Set \(k = 3\sqrt{3}\) and solve:

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

Divide by \(\sqrt{3}\):

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

Expand and simplify:

\(2 + 2\tan^2 \theta = 3 - 9\tan^2 \theta\)

\(11\tan^2 \theta = 1\)

\(\tan^2 \theta = \frac{1}{11}\)

\(\tan \theta = \pm \frac{1}{\sqrt{11}}\)

Find \(\theta\) in the interval \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = 16.8^\circ, 163.2^\circ\)

To solve \(\cos(\theta + 60^\circ) = 2 \sin \theta\), we start by using the angle addition formula for cosine:

\(\cos(\theta + 60^\circ) = \cos \theta \cos 60^\circ - \sin \theta \sin 60^\circ.\)

Substituting the values \(\cos 60^\circ = \frac{1}{2}\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), we get:

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

Rearranging gives:

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

Combine the terms on the right:

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

Divide both sides by \(\sin \theta\) (assuming \(\sin \theta \neq 0\)):

\(\frac{1}{2} \cot \theta = 2 + \frac{\sqrt{3}}{2}.\)

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

\(\cot \theta = 4 + \sqrt{3}.\)

Thus, \(\tan \theta = \frac{1}{4 + \sqrt{3}}.\)

Calculating \(\theta\) gives:

\(\theta = \tan^{-1}\left(\frac{1}{4 + \sqrt{3}}\right).\)

This results in \(\theta \approx 9.9^\circ\) and \(\theta \approx 189.9^\circ\) within the interval \(0^\circ \leq \theta \leq 360^\circ\).

To solve \(\tan(45^\circ - x) = 2 \tan x\), use the tangent subtraction formula:

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

Set this equal to \(2 \tan x\):

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

Cross-multiply to obtain:

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

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

Rearrange to form a quadratic equation:

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

Let \(y = \tan x\). Solve the quadratic equation:

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

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

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

\(y = \frac{-3 \pm \sqrt{9 + 8}}{4}\)

\(y = \frac{-3 \pm \sqrt{17}}{4}\)

Calculate the values:

\(y_1 = \frac{-3 + \sqrt{17}}{4} \approx 0.281\)

\(y_2 = \frac{-3 - \sqrt{17}}{4} \approx -1.781\)

Find \(x\) using \(\tan x = y\):

\(x_1 = \tan^{-1}(0.281) \approx 15.7^\circ\)

\(x_2 = \tan^{-1}(-1.781) \approx 119.3^\circ\)

Thus, the solutions are \(x = 15.7^\circ\) and \(x = 119.3^\circ\).