Question 11
**(b)** Hence solve the equation $\dfrac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta} = \dfrac{5}{\tan\theta}$ for $0° \le \theta \le 180°$.
(In part (a) you proved that $\dfrac{\tan\theta + 7}{\tan^2\theta - 3} \equiv \dfrac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta}$.)
(In part (a) you proved that $\dfrac{\tan\theta + 7}{\tan^2\theta - 3} \equiv \dfrac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta}$.)
