Start with the equation:

\(7 \cos x + 5 = 2 \sin^2 x\)

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

\(7 \cos x + 5 = 2(1 - \cos^2 x)\)

Expand and rearrange:

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

\(2 \cos^2 x + 7 \cos x + 3 = 0\)

Factor the quadratic equation:

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

Set each factor to zero:

\(2 \cos x + 1 = 0\) or \(\cos x + 3 = 0\)

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

Since \(\cos x = -3\) is not possible, solve \(\cos x = -\frac{1}{2}\):

\(x = 120^\circ, 240^\circ\)

(i) Start with the equation \(2 \cos x = 3 \tan x\). Replace \(\tan x\) with \(\frac{\sin x}{\cos x}\), giving \(2 \cos x = 3 \frac{\sin x}{\cos x}\).

Multiply through by \(\cos x\) to eliminate the fraction: \(2 \cos^2 x = 3 \sin x\).

Use the identity \(\cos^2 x = 1 - \sin^2 x\) to rewrite the equation: \(2(1 - \sin^2 x) = 3 \sin x\).

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

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

(ii) Start with the equation \(2 \cos 2y = 3 \tan 2y\). Replace \(\tan 2y\) with \(\frac{\sin 2y}{\cos 2y}\), giving \(2 \cos 2y = 3 \frac{\sin 2y}{\cos 2y}\).

Multiply through by \(\cos 2y\) to eliminate the fraction: \(2 \cos^2 2y = 3 \sin 2y\).

Use the identity \(\cos^2 2y = 1 - \sin^2 2y\) to rewrite the equation: \(2(1 - \sin^2 2y) = 3 \sin 2y\).

Expand and rearrange: \(2 - 2 \sin^2 2y = 3 \sin 2y\).

Rearrange to form a quadratic equation: \(2 \sin^2 2y + 3 \sin 2y - 2 = 0\).

Solve the quadratic equation for \(\sin 2y\): \(\sin 2y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}\).

Calculate the discriminant: \(9 + 16 = 25\).

\(\sin 2y = \frac{-3 \pm 5}{4}\).

\(\sin 2y = \frac{2}{4} = \frac{1}{2}\) or \(\sin 2y = \frac{-8}{4} = -2\) (not possible).

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

Thus, \(y = 15^\circ\) or \(y = 75^\circ\).

(i) Start with the equation \(2 \cos^2 \theta = 3 \sin \theta\). Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), rewrite the equation as:

\(2(1 - \sin^2 \theta) = 3 \sin \theta\)

\(2 - 2 \sin^2 \theta = 3 \sin \theta\)

Rearrange to form a quadratic equation:

\(2 \sin^2 \theta + 3 \sin \theta - 2 = 0\)

Factor the quadratic:

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

Solving gives \(\sin \theta = \frac{1}{2}\) or \(\sin \theta = -2\). Since \(\sin \theta = -2\) is not possible, we have:

\(\theta = 30^\circ\) or \(150^\circ\).

(ii) Given the smallest positive solution is \(10^\circ\), we have \(n \cdot 10^\circ = 30^\circ\), so \(n = 3\).

For the largest solution, consider \(n \cdot \theta = 150^\circ + 360^\circ k\) for integer \(k\). The largest \(\theta\) in \(0^\circ \leq \theta \leq 360^\circ\) is:

\(3 \cdot \theta = 870^\circ\)

\(\theta = 290^\circ\).

(a) Start with the equation \(3 \tan^2 x - 3 \sin^2 x - 4 = 0\).

Replace \(\tan^2 x\) with \(\frac{\sin^2 x}{\cos^2 x}\) and multiply by \(\cos^2 x\):

\(3 \frac{\sin^2 x}{\cos^2 x} - 3 \sin^2 x - 4 = 0\)

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

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

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

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

\(3 \cos^4 x - 10 \cos^2 x + 3 = 0\)

Thus, \(a = 3\), \(b = -10\), \(c = 3\).

(b) Factor the equation:

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

Solving \(3 \cos^2 x - 1 = 0\):

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

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

Solving for \(x\) gives:

\(x = 54.7^\circ, 125.3^\circ\)

(i) Start with the left-hand side (LHS):

\(\tan x + \frac{1}{\tan x} = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\)

Combine the fractions:

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

Using the identity \(\sin^2 x + \cos^2 x = 1\), we have:

\(= \frac{1}{\sin x \cos x}\)

Thus, the identity is proven.

(ii) Given equation:

\(\frac{2}{\sin x \cos x} = 1 + 3 \tan x\)

Using part (i), substitute \(\frac{2}{\sin x \cos x} = 2 \left( \tan x + \frac{1}{\tan x} \right)\):

\(2 \left( \tan x + \frac{1}{\tan x} \right) = 3 \tan x + 1\)

Multiply through by \(\tan x\):

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

Rearrange to form a quadratic equation:

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

Solving the quadratic equation, we find:

\(\tan x = 1\) or \(\tan x = -2\)

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

For \(\tan x = -2\), \(x = 116.6^\circ\).