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 \(x = 2\pi - 1.13 \approx 5.16\).