Explanation

(d)
$\sin{6\theta = 3\sin{2\theta - 4\sin^{3}{2\theta}}}$
$= 3\sin{2\theta - 4.8\cos^{3}{\theta\sin^{3}\theta}}$
$= 3\sin{2\theta - 32\cos^{3}{\theta\sin{\theta(1 - \cos^{2}\theta)}}}$
$\Rightarrow \ \sin{6\theta = 32\cos^{5}{\theta.\sin\theta}} - 32\cos^{3}{\theta\sin{\theta + 3\sin{2\theta}}}$ …(i)
But $\sin{6\theta = 32\cos^{5}{\theta\sin{\theta - 32\cos^{3}{\theta\sin{\theta + 3x}}}}}$ [given]
On comparing Eqs. (i) and (ii), we get
$3x = 3\sin{2\theta\ \ \Rightarrow \ \ \ x = \sin{2\theta}}$