(i) Start with the equation \(3 \sin x \tan x = 8\).

Use the identity \(\tan x = \frac{\sin x}{\cos x}\), so \(3 \sin x \cdot \frac{\sin x}{\cos x} = 8\).

This simplifies to \(\frac{3 \sin^2 x}{\cos x} = 8\).

Use the identity \(\sin^2 x = 1 - \cos^2 x\), so \(\frac{3(1 - \cos^2 x)}{\cos x} = 8\).

Multiply through by \(\cos x\) to clear the fraction: \(3(1 - \cos^2 x) = 8 \cos x\).

Expand and rearrange: \(3 - 3 \cos^2 x = 8 \cos x\).

Rearrange to form a quadratic: \(3 \cos^2 x + 8 \cos x - 3 = 0\).

(ii) Solve the quadratic equation \(3 \cos^2 x + 8 \cos x - 3 = 0\).

Factor the quadratic: \((3 \cos x - 1)(\cos x + 3) = 0\).

Set each factor to zero: \(3 \cos x - 1 = 0\) or \(\cos x + 3 = 0\).

For \(3 \cos x - 1 = 0\), solve \(\cos x = \frac{1}{3}\).

For \(\cos x + 3 = 0\), \(\cos x = -3\) is not possible since \(\cos x\) must be between -1 and 1.

Find \(x\) for \(\cos x = \frac{1}{3}\) in the range \(0^\circ \leq x \leq 360^\circ\).

\(x = 70.5^\circ\) or \(x = 360^\circ - 70.5^\circ = 289.5^\circ\).

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

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

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

Simplify to get:

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

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

Factor the equation:

\(\cos \theta (3 \cos \theta + 2) = 0\)

Set each factor to zero:

1. \(\cos \theta = 0\)

\(\theta = 90^\circ\)

2. \(3 \cos \theta + 2 = 0\)

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

\(\theta = 131.8^\circ\) (to 1 decimal place)

Thus, the solutions are \(\theta = 90^\circ\) or \(\theta = 131.8^\circ\).

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

Divide the entire equation by \(\cos^2 \theta\):

\(\frac{\sin^2 \theta}{\cos^2 \theta} + 3 \frac{\sin \theta \cos \theta}{\cos^2 \theta} = 4\).

This simplifies to \(\tan^2 \theta + 3 \tan \theta = 4\).

Thus, the equation can be written as a quadratic in \(\tan \theta\):

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

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

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

\(\tan \theta = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\).

\(\tan \theta = \frac{-3 \pm \sqrt{9 + 16}}{2}\).

\(\tan \theta = \frac{-3 \pm 5}{2}\).

This gives \(\tan \theta = 1\) or \(\tan \theta = -4\).

For \(\tan \theta = 1\), \(\theta = 45^\circ\).

For \(\tan \theta = -4\), \(\theta = 104^\circ\).

Thus, \(\theta = 45^\circ\) or \(\theta = 104^\circ\).

(i) Start with the given equation:

\(4 \sin^4 \theta + 5 = 7 \cos^2 \theta\)

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

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

\(4 \sin^4 \theta + 5 = 7(1 - \sin^2 \theta)\)

\(4 \sin^4 \theta + 5 = 7 - 7 \sin^2 \theta\)

Let \(x = \sin^2 \theta\), then \(\sin^4 \theta = x^2\).

Substitute \(x\) into the equation:

\(4x^2 + 5 = 7 - 7x\)

Rearrange to form a quadratic equation:

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

(ii) Solve the quadratic equation \(4x^2 + 7x - 2 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 7\), \(c = -2\):

\(x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}\)

\(x = \frac{-7 \pm \sqrt{49 + 32}}{8}\)

\(x = \frac{-7 \pm \sqrt{81}}{8}\)

\(x = \frac{-7 \pm 9}{8}\)

\(x = \frac{2}{8} = \frac{1}{4} \quad \text{or} \quad x = \frac{-16}{8} = -2\)

Since \(x = \sin^2 \theta\), \(x = -2\) is not valid as \(\sin^2 \theta\) must be between 0 and 1.

Thus, \(\sin^2 \theta = \frac{1}{4}\).

\(\sin \theta = \pm \frac{1}{2}\).

For \(\sin \theta = \frac{1}{2}\), \(\theta = 30^\circ, 150^\circ\).

For \(\sin \theta = -\frac{1}{2}\), \(\theta = 210^\circ, 330^\circ\).

Therefore, \(\theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ\).

(a) (i) Expand \((\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2\sin \theta \cos \theta + \sin^2 \theta = 1\).

This simplifies to \(\cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta = 1\).

Since \(\cos^2 \theta + \sin^2 \theta = 1\), we have \(2\sin \theta \cos \theta = 0\).

This implies \(\sin 2\theta = 0\), leading to \(2\theta = 0, \pi, 2\pi\).

Thus, \(\theta = 0, \frac{1}{2}\pi, \pi\).

(ii) Verify \(\cos \theta + \sin \theta = 1\) for \(\theta = 0\) and \(\theta = \frac{1}{2}\pi\).

For \(\theta = 0\), \(\cos 0 + \sin 0 = 1 + 0 = 1\).

For \(\theta = \frac{1}{2}\pi\), \(\cos \frac{1}{2}\pi + \sin \frac{1}{2}\pi = 0 + 1 = 1\).

(b) Prove the identity:

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

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

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

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

Let \(k = \cos \theta + \sin \theta - 1\), then \(\frac{k}{1 - 2 \sin^2 \theta} = 2k\).

\(1 = 2(1 - 2 \sin^2 \theta)\), leading to \(\sin^2 \theta = \frac{1}{4}\).

Thus, \(\sin \theta = \pm \frac{1}{2}\), giving solutions \(\theta = \frac{1}{6}\pi, \frac{5}{6}\pi\).

Including \(\theta = 0\) and \(\theta = \frac{1}{2}\pi\) from part (a)(ii), the solutions are \(0, \frac{1}{6}\pi, \frac{1}{2}\pi, \frac{5}{6}\pi\).