(i) We start with \(\sin x \tan x = \sin x \cdot \frac{\sin x}{\cos x}\).

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

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

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

Thus, \(\sin x \tan x = \frac{1 - \cos^2 x}{\cos x}\).

(ii) Given \(2 \sin x \tan x = 3\), substitute \(\sin x \tan x = \frac{1 - \cos^2 x}{\cos x}\):

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

Multiply through by \(\cos x\):

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

Expand and rearrange:

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

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

Let \(c = \cos x\), then:

\(2c^2 + 3c - 2 = 0\).

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

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

\(c = \frac{-3 \pm \sqrt{9 + 16}}{4}\).

\(c = \frac{-3 \pm 5}{4}\).

\(c = \frac{2}{4} = 0.5\) or \(c = \frac{-8}{4} = -2\).

Since \(\cos x\) must be between -1 and 1, \(\cos x = 0.5\).

Thus, \(x = 60^\circ\) or \(x = 300^\circ\).