(i)[4]
Starting from $\sin(2\theta + \theta)$, expand it to show that $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$.
(ii)[1]
Demonstrate that, once the substitution $x = \frac{2\sin \theta}{\sqrt{3}}$ has been made, the equation $x^3 - x + \frac{1}{6}\sqrt{3} = 0$ may be expressed as $\sin 3\theta = \frac{3}{4}$.
(iii)[4]
Hence solve the equation $x^3 - x + \frac{1}{6}\sqrt{3} = 0$, and give your answers correct to 3 significant figures.