(i) Start with the equation:

\(3 \sin^2 x - 8 \cos x - 7 = 0\).

Using the identity \(\sin^2 x = 1 - \cos^2 x\), substitute:

\(3(1 - \cos^2 x) - 8 \cos x - 7 = 0\).

Simplify to:

\(3 - 3 \cos^2 x - 8 \cos x - 7 = 0\).

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

Multiply through by -1:

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

Factorize:

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

Thus, \(\cos x = -\frac{2}{3}\) or \(\cos x = -2\).

Since \(\cos x = -2\) is not possible for real \(x\), we have \(\cos x = -\frac{2}{3}\).

(ii) Substitute \(\cos(\theta + 70^\circ) = -\frac{2}{3}\) into the equation:

\(\theta + 70^\circ = \cos^{-1}(-\frac{2}{3})\).

Calculate \(\theta + 70^\circ\):

\(\theta + 70^\circ = 131.8^\circ\) or \(228.2^\circ\).

For \(0^\circ \leq \theta \leq 180^\circ\), solve:

\(\theta = 131.8^\circ - 70^\circ = 61.8^\circ\).

\(\theta = 228.2^\circ - 70^\circ = 158.2^\circ\).

Thus, \(\theta = 61.8^\circ\) or \(158.2^\circ\).

(i) Start with the given equation:

\(2 \tan^2 \theta \sin^2 \theta = 1\)

Using the identity \(\tan^2 \theta = \frac{\sin^2 \theta}{1 - \sin^2 \theta}\), substitute:

\(2 \left(\frac{\sin^2 \theta}{1 - \sin^2 \theta}\right) \sin^2 \theta = 1\)

Simplify:

\(\frac{2 \sin^4 \theta}{1 - \sin^2 \theta} = 1\)

Multiply both sides by \(1 - \sin^2 \theta\):

\(2 \sin^4 \theta = 1 - \sin^2 \theta\)

Rearrange to get:

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

(ii) Solve the equation \(2 \sin^4 \theta + \sin^2 \theta - 1 = 0\).

Let \(x = \sin^2 \theta\), then the equation becomes:

\(2x^2 + x - 1 = 0\)

Factorize:

\((2x - 1)(x + 1) = 0\)

So, \(2x - 1 = 0\) or \(x + 1 = 0\).

\(2x - 1 = 0\) gives \(x = \frac{1}{2}\).

\(x + 1 = 0\) gives \(x = -1\), which is not possible since \(x = \sin^2 \theta\) must be non-negative.

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

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

Solutions for \(\theta\) in the range \(0^\circ \leq \theta \leq 360^\circ\) are:

\(\theta = 45^\circ, 135^\circ, 225^\circ, 315^\circ\).

Start with the equation \(15 \sin^2 x = 13 + \cos x\).

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

\(15(1 - \cos^2 x) = 13 + \cos x\).

Expand and simplify:

\(15 - 15 \cos^2 x = 13 + \cos x\).

Rearrange to form a quadratic equation:

\(15 \cos^2 x + \cos x - 2 = 0\).

Factor the quadratic equation:

\((5 \cos x + 2)(3 \cos x - 1) = 0\).

Solve for \(\cos x\):

\(5 \cos x + 2 = 0\) gives \(\cos x = -\frac{2}{5}\).

\(3 \cos x - 1 = 0\) gives \(\cos x = \frac{1}{3}\).

Find the angles \(x\) for each value of \(\cos x\) within the range \(0^\circ \leq x \leq 180^\circ\):

For \(\cos x = -\frac{2}{5}\), \(x \approx 113.6^\circ\).

For \(\cos x = \frac{1}{3}\), \(x \approx 70.5^\circ\).

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

Using \(\tan x = \frac{\sin x}{\cos x}\), rewrite the equation as:

\(2 \sin x \left(\frac{\sin x}{\cos x}\right) + 3 = 0\)

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

Using the identity \(\sin^2 x = 1 - \cos^2 x\), substitute:

\(\frac{2(1 - \cos^2 x)}{\cos x} + 3 = 0\)

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

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

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

Rearrange to get:

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

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

Factor the quadratic:

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

Set each factor to zero:

\(2 \cos x + 1 = 0\) gives \(\cos x = -\frac{1}{2}\)

\(\cos x - 2 = 0\) gives \(\cos x = 2\) (not possible since \(|\cos x| \leq 1\))

For \(\cos x = -\frac{1}{2}\), the solutions in the interval \(0^\circ \leq x \leq 360^\circ\) are:

\(x = 120^\circ\) and \(x = 240^\circ\).

(i) Start with the equation \(2 \tan^2 \theta \cos \theta = 3\).

Replace \(\tan^2 \theta\) with \(\frac{\sin^2 \theta}{\cos^2 \theta}\).

Replace \(\sin^2 \theta\) with \(1 - \cos^2 \theta\).

The equation becomes \(2 \left(\frac{1 - \cos^2 \theta}{\cos^2 \theta}\right) \cos \theta = 3\).

Simplify to get \(2 (1 - \cos^2 \theta) \cos \theta = 3 \cos^2 \theta\).

Rearrange to form \(2 \cos^2 \theta + 3 \cos \theta - 2 = 0\).

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

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

Since \(\cos \theta = -2\) is not possible, solve \(\cos \theta = \frac{1}{2}\).

This gives \(\theta = 60^\circ\) and \(\theta = 300^\circ\) within the range \(0^\circ \leq \theta \leq 360^\circ\).