(i) Use the angle addition formulas: \(\sin(3x + x) = \sin 4x = \sin 3x \cos x + \cos 3x \sin x\) and \(\sin(3x - x) = \sin 2x = \sin 3x \cos x - \cos 3x \sin x\). Adding these gives \(\sin 4x + \sin 2x = 2 \sin 3x \cos x\), hence \(\sin 3x \cos x = \frac{1}{2}(\sin 4x + \sin 2x)\).

(ii) Integrate \(\frac{1}{2}(\sin 4x + \sin 2x)\) from \(x = 0\) to \(x = \frac{1}{2}\pi\):

\(\int_0^{\frac{1}{2}\pi} \frac{1}{2}(\sin 4x + \sin 2x) \, dx = \frac{1}{2} \left[ -\frac{1}{4} \cos 4x - \frac{1}{2} \cos 2x \right]_0^{\frac{1}{2}\pi}\)

\(= \frac{1}{2} \left( \left(-\frac{1}{4} \cos 2\pi - \frac{1}{2} \cos \pi \right) - \left(-\frac{1}{4} \cos 0 - \frac{1}{2} \cos 0 \right) \right)\)

\(= \frac{1}{2} \left( 0 - \left(-\frac{1}{4} - \frac{1}{2} \right) \right) = \frac{1}{2} \left( \frac{3}{4} \right) = \frac{3}{8}\)

Thus, the area \(R = \frac{9}{16}\).

(iii) Differentiate \(y = \sin 3x \cos x = \frac{1}{2}(\sin 4x + \sin 2x)\):

\(\frac{dy}{dx} = 2 \cos 4x + \cos 2x\).

Set \(\frac{dy}{dx} = 0\):

\(2 \cos 4x + \cos 2x = 0\)

Using double angle formulas, solve for \(\cos 2x\):

\(4 \cos^2 2x + \cos 2x - 2 = 0\)

\(\cos 2x = \frac{-1 \pm \sqrt{33}}{8}\)

Calculate \(x\) for \(\cos 2x = 0.593\), giving \(x = 1.29\).