(i) Prove the identity \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 \equiv \frac{1 - \cos x}{1 + \cos x}\).

(ii) Hence solve the equation \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 = \frac{2}{5}\) for \(0 \leq x \leq 2\pi\).