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