(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 \(296.6^\circ\)