Explanation

(b)
We have, $\sin{x - 3\sin{2x + \sin{3x = \cos{x - 3\cos{2x + \cos{3x}}}}}}$
$\Rightarrow \ \sin{x + \sin{3x - 3\sin{2x = \cos{x + \cos{3x - 3\cos{2x}}}}}}$
$\Rightarrow \ 2\sin{2x\cos{x - 3\sin{2x - 2\cos{2x\cos{x + 3\cos{2x = 0}}}}}}$
$\Rightarrow \ \sin{2x(2\cos{x - 3) - \cos{2x(2\cos{x - 3) = 0}}}}$
$\Rightarrow \ (\sin{2x - \cos{2x)(2\cos{x - 3) = 0}}}$
$\Rightarrow \sin{2x = \cos{2x}}\ \ \ \left( \because\cos{x \neq \frac{3}{2}} \right)$
$\Rightarrow 2x = 2n\pi \pm \left( \frac{\pi}{2} - 2x \right)$
Taking +ve sign
$x = \frac{n\pi}{2} + \frac{\pi}{8}$