(a) Prove the identity \(\frac{\tan \theta + 7}{\tan^2 \theta - 3} \equiv \frac{\sin \theta \cos \theta + 7 \cos^2 \theta}{1 - 4 \cos^2 \theta}\).

(b) Hence solve the equation \(\frac{\sin \theta \cos \theta + 7 \cos^2 \theta}{1 - 4 \cos^2 \theta} = \frac{5}{\tan \theta}\) for \(0^\circ \leq \theta \leq 180^\circ\).