(i) Use the formula \(\cos(A + B) = \cos A \cos B - \sin A \sin B\) to expand \(\cos(2x + x)\) as \(\cos 2x \cos x - \sin 2x \sin x\).

Using double angle identities, \(\cos 2x = 2\cos^2 x - 1\) and \(\sin 2x = 2\sin x \cos x\).

Substitute these into the expansion: \((2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x\).

Simplify to get \(2\cos^3 x - \cos x - 2\sin^2 x \cos x\).

Since \(\sin^2 x = 1 - \cos^2 x\), substitute to get \(2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x\).

Simplify to \(4\cos^3 x - 3\cos x\).

(ii) Solve \(4\cos^3 x - 3\cos x + 3\cos x + 1 = 0\) which simplifies to \(4\cos^3 x + 1 = 0\).

\(\cos x = -\frac{1}{2}\) gives \(x = 0.717\pi\) within the interval \(0 \leq x \leq \pi\).

(iii) Use the identity \(\cos^3 x = \frac{1}{4} \left( \cos 3x + 3\cos x \right)\).

Integrate \(\int \cos^3 x \, dx = \frac{1}{12} \sin 3x + \frac{3}{4} \sin x\).

Substitute the limits \(\frac{1}{6}\pi\) and \(\frac{1}{3}\pi\) into the indefinite integral.

Calculate to find \(\frac{1}{24} \left( 9\sqrt{3} - 11 \right)\).