We have,
$\sin 6 \theta=\sin 3(2 \theta)$
$=3 \sin 2 \theta-4 \sin ^3 2 \theta$
$=3 \cdot 2 \sin \theta \cos \theta-4 \cdot 8 \cdot \cos ^3 \theta \sin ^3 \theta$
$=6 \sin \theta \cos \theta-32 \cos ^3 \theta \sin \theta\left(1-\cos ^2 \theta\right)$
$\sin 6 \theta=32 \cos ^5 \theta \cdot \sin \theta$ $-32 \cos ^3 \theta \sin \theta+3 \sin 2 \theta \ldots$ (i)
Given that,
$\sin 6 \theta=32 \cos ^5 \theta \sin \theta-32 \cos ^3 \theta \sin \theta+3 x$ $\ldots$ (ii)
On comparing Eqs. (i) and (ii), we get
$3 x=3 \sin 2 \theta \Rightarrow x=\sin 2 \theta$