Explanation

(a)
$\because\ \sin{3x = 3\sin{x - 4\sin^{3}x}}$
$\therefore\ \sin^{3}{x = \frac{1}{4}(3\sin{x - \sin{3x)}}}$
and $\cos{3x = 4\cos^{3}{x - 3\cos x}}$
$\Rightarrow \ \cos^{3}{x = \frac{1}{4}(\cos{3x + 3\cos{x)}}}$
$\therefore\ \cos{3x\cos^{3}{x + \sin{3x\sin^{3}x}}}$
$= \frac{1}{4}\lbrack\cos^{2}{3x + 3\cos{x\cos{3x + 3\sin{x\sin{3x} - \sin^{2}{3x\rbrack}}}}}$
$= \frac{1}{4}\lbrack 3\cos{2x + \cos{6x\rbrack = \cos^{3}{2x}}}$
$\Rightarrow \ \cos{2x = 0\ \Rightarrow}\ 2x = (2n + 1)\frac{\pi}{2}$
$\Rightarrow \ x = (2n + 1)\frac{\pi}{4}$