Show that the equation

$$\frac{2\sin \theta \cos \theta }{3\sin \theta -2}=3\tan \theta ,\sin \theta \ne \frac{2}{3}$$

can be written in the form

$$2{\sin }^3\theta +9{\sin }^2\theta -8\sin \theta =0$$

Hence solve, for $-\frac{\pi }{2}<x<\frac{\pi }{2}$

$$\frac{2\sin x\cos x}{3\sin x-2}=3\tan x$$

giving your answers to 3 decimal places where appropriate.