(i) Start with the equation:

\(\frac{\cos \theta - 4}{\sin \theta} - \frac{4 \sin \theta}{5 \cos \theta - 2} = 0\)

Multiply through by \(\sin \theta (5 \cos \theta - 2)\) to clear the denominators:

\((\cos \theta - 4)(5 \cos \theta - 2) - 4 \sin^2 \theta = 0\)

Expand and simplify:

\(5 \cos^2 \theta - 22 \cos \theta + 8 - 4(1 - \cos^2 \theta) = 0\)

\(5 \cos^2 \theta - 22 \cos \theta + 8 - 4 + 4 \cos^2 \theta = 0\)

Combine like terms:

\(9 \cos^2 \theta - 22 \cos \theta + 4 = 0\)

Thus, the equation is shown as required.

(ii) Solve the quadratic equation:

\(9 \cos^2 \theta - 22 \cos \theta + 4 = 0\)

Using the quadratic formula \(\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9\), \(b = -22\), \(c = 4\):

\(\cos \theta = \frac{22 \pm \sqrt{(-22)^2 - 4 \times 9 \times 4}}{18}\)

\(\cos \theta = \frac{22 \pm \sqrt{484 - 144}}{18}\)

\(\cos \theta = \frac{22 \pm \sqrt{340}}{18}\)

\(\cos \theta = \frac{22 \pm 18.44}{18}\)

\(\cos \theta = 0.1978 \text{ or } 2.247\)

Since \(\cos \theta\) must be between -1 and 1, only \(\cos \theta = 0.1978\) is valid.

Find \(\theta\):

\(\theta = \cos^{-1}(0.1978) \approx 78.6^\circ\)

For the second solution in the range \(0^\circ \leq \theta \leq 360^\circ\), use \(360^\circ - 78.6^\circ = 281.4^\circ\).

Thus, \(\theta = 78.6^\circ, 281.4^\circ\).

Start by multiplying both sides by the denominator \(3 + 2 \tan x\) to eliminate the fraction:

\(5 + 2 \tan x = (1 + \tan x)(3 + 2 \tan x)\)

Expand the right side:

\(5 + 2 \tan x = 3 + 3 \tan x + 2 \tan x + 2 \tan^2 x\)

Combine like terms:

\(5 + 2 \tan x = 3 + 5 \tan x + 2 \tan^2 x\)

Rearrange to form a quadratic equation:

\(2 \tan^2 x + 3 \tan x - 2 = 0\)

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

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

\(\tan x = \frac{-3 \pm \sqrt{9 + 16}}{4}\)

\(\tan x = \frac{-3 \pm 5}{4}\)

\(\tan x = \frac{2}{4} = 0.5\) or \(\tan x = \frac{-8}{4} = -2\)

For \(\tan x = 0.5\), \(x = \tan^{-1}(0.5) \approx 0.464\) (accept \(0.148\pi\))

For \(\tan x = -2\), \(x = \tan^{-1}(-2) \approx 2.03\) (accept \(0.648\pi\))

(i) Start with the equation:

\(\frac{\cos \theta + 4}{\sin \theta + 1} + 5 \sin \theta - 5 = 0\)

Multiply throughout by \(\sin \theta + 1\):

\(\cos \theta + 4 + 5 \sin \theta (\sin \theta + 1) - 5(\sin \theta + 1) = 0\)

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

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

\(\cos \theta - 1 + 5 \sin^2 \theta = 0\)

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

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

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

\(5 \cos^2 \theta - \cos \theta - 4 = 0\)

(ii) Solve \(5 \cos^2 \theta - \cos \theta - 4 = 0\).

Factorize or use the quadratic formula:

\(\cos \theta = 1 \quad \text{and} \quad \cos \theta = -0.8\)

For \(\cos \theta = 1\), \(\theta = 0^\circ, 360^\circ\).

For \(\cos \theta = -0.8\), \(\theta = 143.1^\circ, 216.9^\circ\).

Thus, \(\theta = [0^\circ, 360^\circ], [143.1^\circ], [216.9^\circ]\).

(i) Start with the equation \(\cos 2x(\tan^2 2x + 3) + 3 = 0\).

Use the identity \(\tan^2 2x = \frac{\sin^2 2x}{\cos^2 2x}\).

Replace \(\sin^2 2x\) with \(1 - \cos^2 2x\):

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

Substitute back into the equation:

\(\cos 2x((\sec^2 2x - 1) + 3) + 3 = 0\).

Simplify to get:

\(\cos 2x(\sec^2 2x + 2) + 3 = 0\).

Multiply through by \(\cos^2 2x\) and rearrange:

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

(ii) Solve \(2 \cos^2 2x + 3 \cos 2x + 1 = 0\).

Let \(y = \cos 2x\), then the equation becomes:

\(2y^2 + 3y + 1 = 0\).

Factorize to find \(y = -\frac{1}{2}\) or \(y = -1\).

For \(\cos 2x = -\frac{1}{2}\), \(2x = 120^\circ\) or \(240^\circ\), so \(x = 60^\circ\) or \(120^\circ\).

For \(\cos 2x = -1\), \(2x = 180^\circ\), so \(x = 90^\circ\).

Thus, the solutions are \(x = 60^\circ, 90^\circ, 120^\circ\).

(i) Expand the left-hand side: \((\sin \theta + 2 \cos \theta)(1 + \sin \theta - \cos \theta) = \sin \theta + \sin^2 \theta - \sin \theta \cos \theta + 2 \cos \theta + 2 \cos \theta \sin \theta - 2 \cos^2 \theta\).

Rearrange terms: \(\sin \theta + \sin^2 \theta - \sin \theta \cos \theta + 2 \cos \theta + 2 \cos \theta \sin \theta - 2 \cos^2 \theta = \sin \theta + 2 \cos \theta + \sin^2 \theta + \sin \theta \cos \theta - 2 \cos^2 \theta\).

Use \(\sin^2 \theta = 1 - \cos^2 \theta\): \(\sin \theta + 2 \cos \theta + (1 - \cos^2 \theta) + \sin \theta \cos \theta - 2 \cos^2 \theta = \sin \theta + 2 \cos \theta + 1 - \cos^2 \theta + \sin \theta \cos \theta - 2 \cos^2 \theta\).

Simplify: \(1 + \sin \theta + 2 \cos \theta - 3 \cos^2 \theta + \sin \theta \cos \theta = 0\).

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

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

Let \(x = \cos \theta\), then \(3x^2 - 2x - 1 = 0\).

Factorize: \((3x + 1)(x - 1) = 0\).

Solutions: \(x = 1\) or \(x = -\frac{1}{3}\).

For \(\cos \theta = 1\), \(\theta = 0^\circ\).

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