(i) Start with the left-hand side of the identity:

\(\csc 2\theta + \cot 2\theta = \frac{1}{\sin 2\theta} + \frac{\cos 2\theta}{\sin 2\theta}\)

Combine the terms:

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

Using the double angle identity \(\cos 2\theta = 1 - 2\sin^2 \theta\), substitute:

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

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

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

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

\(= \frac{\cos \theta}{\sin \theta}\)

\(= \cot \theta\)

Thus, the identity is proven.

(ii) Given \(\csc 2\theta + \cot 2\theta = 2\), use the proven identity:

\(\cot \theta = 2\)

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

Calculate \(\theta\):

\(\theta = 26.6^\circ\)

Since \(\cot \theta\) is periodic with period \(180^\circ\), the other solution is:

\(\theta = 26.6^\circ + 180^\circ = 206.6^\circ\)

Thus, \(\theta = 26.6^\circ, 206.6^\circ\).

To solve the equation \(\tan x \tan 2x = 1\), we start by using the identity for \(\tan 2x\):

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

Substitute this into the equation:

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

Simplify the equation:

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

Cross-multiply to clear the fraction:

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

Rearrange the terms:

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

Divide by 3:

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

Take the square root:

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

Thus, \(x = 30^\circ, 150^\circ\) in the given interval \(0^\circ < x < 180^\circ\).

Part (i):

We need to prove the identity \(\cos 4\theta + 4\cos 2\theta \equiv 8\cos^4 \theta - 3\).

Using the double angle formulas:

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

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

Expanding \(\cos 4\theta\):

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

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

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

Now, substitute \(\cos 4\theta\) and \(\cos 2\theta\) into the left-hand side:

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

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

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

Thus, the identity is proven.

Part (ii):

We need to solve \(\cos 4\theta + 4\cos 2\theta = 2\).

Using the identity from part (i):

\(8\cos^4 \theta - 3 = 2\)

\(8\cos^4 \theta = 5\)

\(\cos^4 \theta = \frac{5}{8}\)

\(\cos^2 \theta = \sqrt{\frac{5}{8}}\)

\(\cos \theta = \pm \sqrt{\frac{5}{8}}\)

Calculating \(\theta\) for \(\cos \theta = \sqrt{\frac{5}{8}}\):

\(\theta = \cos^{-1}(\sqrt{\frac{5}{8}}) \approx 27.3^\circ\)

Other solutions in the range \(0^\circ \leq \theta \leq 360^\circ\) are:

\(360^\circ - 27.3^\circ = 332.8^\circ\)

For \(\cos \theta = -\sqrt{\frac{5}{8}}\):

\(\theta = 180^\circ - 27.3^\circ = 152.8^\circ\)

\(180^\circ + 27.3^\circ = 207.2^\circ\)

Thus, \(\theta = 27.3^\circ, 152.8^\circ, 207.2^\circ, 332.8^\circ\).

To solve the equation \(\cos \theta + 3 \cos 2\theta = 2\), we start by using the double angle formula for cosine: \(\cos 2\theta = 2\cos^2 \theta - 1\).

Substitute \(\cos 2\theta\) in the equation:

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

Simplify the equation:

\(\cos \theta + 6\cos^2 \theta - 3 = 2\)

Rearrange terms:

\(6\cos^2 \theta + \cos \theta - 5 = 0\)

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

\(6x^2 + x - 5 = 0\)

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

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

\(x = \frac{-1 \pm \sqrt{1 + 120}}{12}\)

\(x = \frac{-1 \pm \sqrt{121}}{12}\)

\(x = \frac{-1 \pm 11}{12}\)

\(x = \frac{10}{12} = \frac{5}{6} \quad \text{or} \quad x = \frac{-12}{12} = -1\)

Thus, \(\cos \theta = \frac{5}{6}\) or \(\cos \theta = -1\).

For \(\cos \theta = \frac{5}{6}\):

\(\theta = \cos^{-1} \left( \frac{5}{6} \right) \approx 33.6^\circ\)

For \(\cos \theta = -1\):

\(\theta = 180^\circ\)

Therefore, the solutions are \(\theta = 33.6^\circ\) and \(\theta = 180^\circ\).

(a) Start with the given equation:

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

Using double angle identities, we have:

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

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

Substitute these into the equation:

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

Rearrange terms:

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

(b) Factor the equation:

\((\cos \theta - \sin \theta)(\cos \theta + 3 \sin \theta) = 0\)

Set each factor to zero:

\(\cos \theta - \sin \theta = 0\)

\(\tan \theta = 1\)

\(\theta = 45^\circ\)

And:

\(\cos \theta + 3 \sin \theta = 0\)

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

\(\theta \approx 161.6^\circ\)

Thus, the solutions are \(\theta = 45^\circ\) and \(\theta \approx 161.6^\circ\).