(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.