(i) Start with the equation \(\frac{4 \cos \theta}{\tan \theta} + 15 = 0\).

Replace \(\tan \theta\) with \(\frac{\sin \theta}{\cos \theta}\), giving \(\frac{4 \cos \theta}{\frac{\sin \theta}{\cos \theta}} + 15 = 0\).

This simplifies to \(\frac{4 \cos^2 \theta}{\sin \theta} + 15 = 0\).

Multiply through by \(\sin \theta\) to clear the fraction: \(4 \cos^2 \theta + 15 \sin \theta = 0\).

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to get \(4(1 - \sin^2 \theta) + 15 \sin \theta = 0\).

Expand and rearrange: \(4 - 4 \sin^2 \theta + 15 \sin \theta = 0\).

Rearrange to \(4 \sin^2 \theta - 15 \sin \theta - 4 = 0\), as required.

(ii) Solve \(4 \sin^2 \theta - 15 \sin \theta - 4 = 0\).

Let \(s = \sin \theta\). The equation becomes \(4s^2 - 15s - 4 = 0\).

Use the quadratic formula: \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -15\), \(c = -4\).

Calculate the discriminant: \(b^2 - 4ac = (-15)^2 - 4 \times 4 \times (-4) = 225 + 64 = 289\).

\(s = \frac{15 \pm \sqrt{289}}{8} = \frac{15 \pm 17}{8}\).

This gives \(s = 4\) or \(s = -\frac{1}{4}\).

Since \(\sin \theta\) must be between -1 and 1, \(s = 4\) is not valid.

Thus, \(\sin \theta = -\frac{1}{4}\).

Find \(\theta\) using \(\sin^{-1}(-\frac{1}{4})\), which gives \(\theta = 194.5^\circ\) or \(\theta = 345.5^\circ\) within the range \(0^\circ \leq \theta \leq 360^\circ\).

(i) Start with \(\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta}\).

Divide the numerator and the denominator by \(\cos \theta\):

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

This proves the identity.

(ii) Using the identity from part (i), \(\frac{\tan \theta - 1}{\tan \theta + 1} = \frac{\tan \theta}{6}\).

Let \(t = \tan \theta\), then \(\frac{t - 1}{t + 1} = \frac{t}{6}\).

Cross-multiply to get:

\(6(t - 1) = t(t + 1)\).

Expand and simplify:

\(6t - 6 = t^2 + t\).

Rearrange to form a quadratic equation:

\(t^2 - 5t + 6 = 0\).

Factorize the quadratic:

\((t - 2)(t - 3) = 0\).

So, \(t = 2\) or \(t = 3\).

Thus, \(\tan \theta = 2\) or \(\tan \theta = 3\).

Find \(\theta\) using inverse tangent:

\(\theta = \tan^{-1}(2) \approx 63.4^\circ\) or \(\theta = \tan^{-1}(3) \approx 71.6^\circ\).

(i) Start with the equation \(1 + \sin x \tan x = 5 \cos x\).

Replace \(\tan x\) with \(\frac{\sin x}{\cos x}\):

\(1 + \frac{\sin^2 x}{\cos x} = 5 \cos x\)

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

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

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

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

Rearrange to form a quadratic equation:

\(6 \cos^2 x - \cos x - 1 = 0\)

(ii) Solve the quadratic equation \(6 \cos^2 x - \cos x - 1 = 0\).

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

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

\(\cos x = \frac{1 \pm \sqrt{1 + 24}}{12}\)

\(\cos x = \frac{1 \pm 5}{12}\)

\(\cos x = \frac{6}{12} = \frac{1}{2}\) or \(\cos x = \frac{-4}{12} = -\frac{1}{3}\)

For \(\cos x = \frac{1}{2}\), \(x = 60^\circ\).

For \(\cos x = -\frac{1}{3}\), \(x \approx 109.5^\circ\).

(i) Start with the equation \(4 \sin^2 x + 8 \cos x - 7 = 0\). Use the identity \(\sin^2 x = 1 - \cos^2 x\) to rewrite the equation as:

\(4(1 - \cos^2 x) + 8 \cos x - 7 = 0\)

Simplify to get:

\(4 - 4 \cos^2 x + 8 \cos x - 7 = 0\)

\(-4 \cos^2 x + 8 \cos x - 3 = 0\)

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

\(4c^2 - 8c + 3 = 0\)

Factor the quadratic equation:

\((2c - 1)(2c - 3) = 0\)

Thus, \(c = \frac{1}{2}\) or \(c = \frac{3}{2}\). Since \(\cos x\) must be between -1 and 1, \(c = \frac{3}{2}\) is not valid.

For \(c = \frac{1}{2}\), \(\cos x = \frac{1}{2}\) gives \(x = 60^\circ\) or \(x = 300^\circ\).

(ii) For \(4 \sin^2 \left(\frac{1}{2} \theta\right) + 8 \cos \left(\frac{1}{2} \theta\right) - 7 = 0\), let \(x = \frac{1}{2} \theta\).

From part (i), \(x = 60^\circ\) or \(x = 300^\circ\).

Thus, \(\frac{1}{2} \theta = 60^\circ\) or \(\frac{1}{2} \theta = 300^\circ\).

Therefore, \(\theta = 120^\circ\) or \(\theta = 600^\circ\). Since \(\theta\) must be between \(0^\circ\) and \(360^\circ\), \(\theta = 120^\circ\) is the only valid solution.

(i) Start with the equation \(2 \cos^2 \theta = \tan^2 \theta\).

We know \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\).

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

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have \(\sin^2 \theta = 1 - \cos^2 \theta\).

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

Rearrange to form a quadratic: \(2c^4 - c^2 - 1 = 0\), where \(c = \cos \theta\).

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

This gives \(c^2 = \frac{1}{2}\) or \(c^2 = -1\).

Since \(c^2 = -1\) has no real solutions, we consider \(c^2 = \frac{1}{2}\).

Thus, \(c = \pm \frac{1}{\sqrt{2}}\).

For \(c = \frac{1}{\sqrt{2}}\), \(\theta = \frac{\pi}{4}\).

For \(c = -\frac{1}{\sqrt{2}}\), \(\theta = \frac{3\pi}{4}\).

Therefore, the solutions are \(\theta = \frac{\pi}{4}\) or \(\frac{3\pi}{4}\).