By first expanding \(\sin ( 2 \theta + \theta )\), show that
$$\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$$
Show that, after making the substitution \(x = \frac { 2 \sin \theta } { \sqrt { 3 } }\), the equation \(x ^ { 3 } - x + \frac { 1 } { 6 } \sqrt { } 3 = 0\) can be written in the form \(\sin 3 \theta = \frac { 3 } { 4 }\).
Hence solve the equation
$$x ^ { 3 } - x + \frac { 1 } { 6 } \sqrt { } 3 = 0$$
giving your answers correct to 3 significant figures.