Problem

(a) Show that the equation

$$6\sin\theta+\frac{1}{\tan\theta}=\frac{4}{\sin\theta}$$

can be written in the form

$$6\cos^2\theta-\cos\theta-2=0.$$

(b) Hence, solve the equation

$$6\sin\theta+\frac{1}{\tan\theta}=\frac{4}{\sin\theta}$$

for $0^\circ\leq \theta \leq 360^\circ$.