(i) Start with the equation \(3 \cos^4 \theta + 4 \sin^2 \theta - 3 = 0\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\).

Substitute \(\sin^2 \theta = 1 - x\) where \(x = \cos^2 \theta\).

The equation becomes \(3x^2 + 4(1-x) - 3 = 0\).

Simplify to get \(3x^2 - 4x + 1 = 0\).

(ii) Solve the quadratic equation \(3x^2 - 4x + 1 = 0\).

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

Calculate the discriminant: \((-4)^2 - 4 \times 3 \times 1 = 4\).

\(x = \frac{4 \pm 2}{6}\), giving \(x = 1\) or \(x = \frac{1}{3}\).

For \(x = 1\), \(\cos^2 \theta = 1\), so \(\cos \theta = \pm 1\).

\(\theta = 0^\circ, 180^\circ\).

For \(x = \frac{1}{3}\), \(\cos^2 \theta = \frac{1}{3}\), so \(\cos \theta = \pm \frac{1}{\sqrt{3}}\).

\(\theta = 54.7^\circ, 125.3^\circ\).

(a) Expand \((2x - 1)(4x^2 + 2x - 1)\):

\((2x - 1)(4x^2 + 2x - 1) = 8x^3 + 4x^2 - 2x - 4x^2 - 2x + 1 = 8x^3 - 4x + 1\).

Thus, the identity is verified.

(b) Start with the left-hand side:

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

Thus, the identity is proven.

(c) Using the results of (a) and (b), solve:

\(\frac{1}{1 - 2\cos^2 \theta} = 4\cos \theta\).

Cross-multiply to get:

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

Simplify to:

\(4\cos \theta - 8\cos^3 \theta = 1\).

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

\(8x^3 - 4x + 1 = 0\).

Using the identity from (a), solve for \(x\):

\((2x - 1)(4x^2 + 2x - 1) = 0\).

\(2x - 1 = 0\) gives \(x = \frac{1}{2}\), \(\cos \theta = \frac{1}{2}\), \(\theta = 60^\circ\).

\(4x^2 + 2x - 1 = 0\) gives \(x = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm 4}{8}\).

\(x = \frac{1}{4}\), \(\cos \theta = \frac{1}{4}\), \(\theta = 75.5^\circ\) (not in range).

\(x = -\frac{3}{4}\), \(\cos \theta = -\frac{3}{4}\), \(\theta = 144^\circ\).

Thus, \(\theta = 60^\circ, 72^\circ, 144^\circ\).

(i) Start with the equation:

\(4 \tan x + 3 \cos x + \frac{1}{\cos x} = 0\)

Multiply through by \(\cos x\) to eliminate the fraction:

\(4 \sin x + 3 \cos^2 x + 1 = 0\)

Use the identity \(\cos^2 x = 1 - \sin^2 x\):

\(4 \sin x + 3(1 - \sin^2 x) + 1 = 0\)

Simplify to form a quadratic in \(\sin x\):

\(3 \sin^2 x - 4 \sin x - 4 = 0\)

Solving this quadratic gives \(\sin x = -\frac{2}{3}\).

(ii) Substitute \(\sin(2x - 20^\circ) = -\frac{2}{3}\) into the equation:

\(2x - 20^\circ = 221.8^\circ, 318.2^\circ\)

Solve for \(x\):

\(x = 120.9^\circ, 169.1^\circ\)

Start with the equation:

\(3 \sin^2 2\theta + 8 \cos 2\theta = 0\)

Use the identity \(\sin^2 2\theta = 1 - \cos^2 2\theta\) to rewrite the equation:

\(3(1 - \cos^2 2\theta) + 8 \cos 2\theta = 0\)

Simplify to:

\(3 - 3\cos^2 2\theta + 8 \cos 2\theta = 0\)

Rearrange to form a quadratic equation in \(\cos 2\theta\):

\(3\cos^2 2\theta - 8 \cos 2\theta - 3 = 0\)

Let \(x = \cos 2\theta\), then solve the quadratic:

\(3x^2 - 8x - 3 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -8\), \(c = -3\):

\(x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times (-3)}}{2 \times 3}\)

\(x = \frac{8 \pm \sqrt{64 + 36}}{6}\)

\(x = \frac{8 \pm \sqrt{100}}{6}\)

\(x = \frac{8 \pm 10}{6}\)

\(x = 3.0 \text{ or } x = -\frac{1}{3}\)

Since \(\cos 2\theta\) must be between -1 and 1, we take \(\cos 2\theta = -\frac{1}{3}\).

Find \(2\theta\):

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

\(2\theta \approx 109.47^\circ \text{ or } 250.53^\circ\)

Thus, \(\theta \approx 54.7^\circ \text{ or } 125.3^\circ\).

(i) Start with the expression:

\(\frac{\tan \theta + 1}{1 + \cos \theta} + \frac{\tan \theta - 1}{1 - \cos \theta}\)

Combine the fractions:

\(\frac{(\tan \theta + 1)(1 - \cos \theta) + (\tan \theta - 1)(1 + \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)}\)

Simplify the numerator:

\(\tan \theta - \tan \theta \cos \theta + 1 - \cos \theta + \tan \theta + \tan \theta \cos \theta - 1 - \cos \theta\)

\(= 2\tan \theta - 2\cos \theta\)

The denominator is:

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

Thus, the expression simplifies to:

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

(ii) Set the expression to zero:

\(\frac{2(\tan \theta - \cos \theta)}{\sin^2 \theta} = 0\)

This implies:

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

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

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

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

\(\sin^2 \theta + \sin \theta - 1 = 0\)

Using the quadratic formula, solve for \(\sin \theta\):

\(\sin \theta = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\)

\(\sin \theta = 0.618\)

\(\theta = 38.2^\circ\)