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

(ii) Hence solve the equation \(\left( \frac{1}{\cos 2x} - \tan 2x \right)^2 = \frac{1}{3}\) for \(0 \leq x \leq \pi\).