(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:

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