(i) Start with the identity \(\cos^2 x = 1 - \sin^2 x\).

Then, \(\cos^4 x = (\cos^2 x)^2 = (1 - \sin^2 x)^2\).

Expanding, \((1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x\).

Thus, \(\cos^4 x = 1 - 2\sin^2 x + \sin^4 x\).

(ii) Substitute \(\cos^4 x = 1 - 2\sin^2 x + \sin^4 x\) into the equation:

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

Simplify to get \(9\sin^4 x + 1 - 2\sin^2 x = 2 - 2\sin^2 x\).

Thus, \(9\sin^4 x = 1\).

\(\sin^4 x = \frac{1}{9}\).

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

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

Solving for \(x\) gives \(x = 35.3^\circ, 144.7^\circ, 215.3^\circ, 324.7^\circ\).

(i) Start with the equation \(3 \sin x \tan x - \cos x + 1 = 0\). Using \(\tan x = \frac{\sin x}{\cos x}\), rewrite the equation as \(3 \sin^2 x - \cos^2 x + \cos x = 0\). Substitute \(\sin^2 x = 1 - \cos^2 x\) to get \(3(1 - \cos^2 x) - \cos^2 x + \cos x = 0\). Simplify to \(3 - 3\cos^2 x - \cos^2 x + \cos x = 0\), which becomes \(4\cos^2 x - \cos x - 3 = 0\). Solve the quadratic equation \(4c^2 - c - 3 = 0\) for \(c = \cos x\). The solutions are \(\cos x = -\frac{3}{4}\) and \(\cos x = 1\). For \(\cos x = -\frac{3}{4}\), \(x = 2.42\) (allow \(0.77\pi\)). For \(\cos x = 1\), \(x = 0\).

(ii) The equation is \(3 \sin 2x \tan 2x - \cos 2x + 1 = 0\). Using \(\tan 2x = \frac{\sin 2x}{\cos 2x}\), rewrite as \(3 \sin^2 2x - \cos^2 2x + \cos 2x = 0\). Substitute \(\sin^2 2x = 1 - \cos^2 2x\) to get \(3(1 - \cos^2 2x) - \cos^2 2x + \cos 2x = 0\). Simplify to \(4\cos^2 2x - \cos 2x - 3 = 0\). Solve the quadratic equation \(4c^2 - c - 3 = 0\) for \(c = \cos 2x\). The solutions are \(2x = 2\pi - 2.42\) or \(360 - 138.6\). Thus, \(x = 1.21\) (\(0.385\pi\)), \(1.93\) (\(0.614\pi\)), \(0\), \(\pi\) (\(3.14\)).

(a) Start with the equation:

\(4 \sin x + \frac{5}{\tan x} + \frac{2}{\sin x} = 0\)

Multiply through by \(\sin x\) to eliminate the fractions:

\(4 \sin^2 x + 5 \cos x + 2 = 0\)

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

\(4(1 - \cos^2 x) + 5 \cos x + 2 = 0\)

Simplify to get:

\(4 - 4 \cos^2 x + 5 \cos x + 2 = 0\)

Rearrange to form a quadratic in \(\cos x\):

\(4 \cos^2 x - 5 \cos x - 6 = 0\)

Thus, \(a = 4, b = -5, c = -6\).

(b) Solve the quadratic equation:

\(4 \cos^2 x - 5 \cos x - 6 = 0\)

Factorize:

\((4 \cos x + 3)(\cos x - 2) = 0\)

Set each factor to zero:

\(4 \cos x + 3 = 0\) or \(\cos x - 2 = 0\)

\(\cos x = -\frac{3}{4}\) or \(\cos x = 2\)

Since \(\cos x = 2\) is not possible, solve \(\cos x = -\frac{3}{4}\):

\(x = \cos^{-1}(-\frac{3}{4})\)

\(x \approx 138.6^\circ\) and \(x \approx 221.4^\circ\)

Start with the equation \(3 \sin^2 \theta = 4 \cos \theta - 1\).

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

\(3(1 - \cos^2 \theta) = 4 \cos \theta - 1\)

Expand and simplify:

\(3 - 3 \cos^2 \theta = 4 \cos \theta - 1\)

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

\(3 \cos^2 \theta + 4 \cos \theta - 4 = 0\)

Let \(c = \cos \theta\). The equation becomes:

\(3c^2 + 4c - 4 = 0\)

Use the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 3\), \(b = 4\), \(c = -4\):

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

\(c = \frac{-4 \pm \sqrt{16 + 48}}{6}\)

\(c = \frac{-4 \pm \sqrt{64}}{6}\)

\(c = \frac{-4 \pm 8}{6}\)

Solutions for \(c\) are \(c = \frac{2}{3}\) or \(c = -2\).

Since \(\cos \theta\) must be between -1 and 1, \(c = -2\) is not valid.

Thus, \(\cos \theta = \frac{2}{3}\).

Find \(\theta\) using \(\cos^{-1}\):

\(\theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ\)

The second solution in the range \(0^\circ \leq \theta \leq 360^\circ\) is:

\(\theta = 360^\circ - 48.2^\circ = 311.8^\circ\)

Therefore, \(\theta = 48.2^\circ\) or \(\theta = 311.8^\circ\).

Start with the equation:

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

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

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

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

\(1 + 3(1 - \cos^2 \theta) + 4 \cos \theta = 0\)

Simplify:

\(1 + 3 - 3 \cos^2 \theta + 4 \cos \theta = 0\)

\(3 \cos^2 \theta - 4 \cos \theta - 4 = 0\)

Now solve \(3 \cos^2 \theta - 4 \cos \theta - 4 = 0\):

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

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

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

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

\(x = \frac{4 \pm \sqrt{16 + 48}}{6}\)

\(x = \frac{4 \pm \sqrt{64}}{6}\)

\(x = \frac{4 \pm 8}{6}\)

\(x = 2 \text{ or } x = -\frac{2}{3}\)

Since \(\cos \theta\) cannot be 2, use \(\cos \theta = -\frac{2}{3}\):

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

\(\theta \approx 131.8^\circ \text{ or } 228.2^\circ\)