- (a) Show that the equation
$$\frac { 3 \sin \theta \cos \theta } { 2 \sin \theta - 1 } = 5 \tan \theta \quad \sin \theta \neq \frac { 1 } { 2 }$$
can be written in the form
$$3 \sin ^ { 3 } \theta + 10 \sin ^ { 2 } \theta - 8 \sin \theta = 0$$
(b) Hence solve, for \(- \frac { \pi } { 4 } < x < \frac { \pi } { 4 }\)
$$\frac { 3 \sin 2 x \cos 2 x } { 2 \sin 2 x - 1 } = 5 \tan 2 x$$
giving your answers to 3 decimal places where appropriate.