(i) Expand \(\sin(2\theta + \theta)\) using the angle addition formula:

\(\sin(2\theta + \theta) = \sin 2\theta \cos \theta + \cos 2\theta \sin \theta\)

Using double angle identities:

\(\sin 2\theta = 2 \sin \theta \cos \theta\)

\(\cos 2\theta = 1 - 2 \sin^2 \theta\)

Substitute these into the expansion:

\(\sin 3\theta = (2 \sin \theta \cos \theta) \cos \theta + (1 - 2 \sin^2 \theta) \sin \theta\)

\(= 2 \sin \theta \cos^2 \theta + \sin \theta - 2 \sin^3 \theta\)

\(= \sin \theta (2 \cos^2 \theta + 1 - 2 \sin^2 \theta)\)

\(= \sin \theta (2(1 - \sin^2 \theta) + 1 - 2 \sin^2 \theta)\)

\(= \sin \theta (3 - 4 \sin^2 \theta)\)

\(= 3 \sin \theta - 4 \sin^3 \theta\)

This proves the identity.

(ii) Substitute \(x = \frac{2 \sin \theta}{\sqrt{3}}\) into the equation:

\(x^3 - x + \frac{1}{6}\sqrt{3} = 0\)

\(\left(\frac{2 \sin \theta}{\sqrt{3}}\right)^3 - \frac{2 \sin \theta}{\sqrt{3}} + \frac{1}{6}\sqrt{3} = 0\)

\(\frac{8 \sin^3 \theta}{3\sqrt{3}} - \frac{2 \sin \theta}{\sqrt{3}} + \frac{1}{6}\sqrt{3} = 0\)

Multiply through by \(3\sqrt{3}\):

\(8 \sin^3 \theta - 6 \sin \theta + \frac{1}{2} \cdot 3 = 0\)

\(8 \sin^3 \theta - 6 \sin \theta + \frac{3}{2} = 0\)

\(8 \sin^3 \theta - 6 \sin \theta = -\frac{3}{2}\)

\(\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta = \frac{3}{4}\)

This confirms the form \(\sin 3\theta = \frac{3}{4}\).

(iii) Solve \(x^3 - x + \frac{1}{6}\sqrt{3} = 0\):

Using numerical methods or a calculator, find:

\(x = 0.322, 0.799, -1.12\)