If $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ then $x = $
If $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ then $x = $
- A
$\cos \theta $
- B
$\cos 2\theta $
- C
$\sin \theta $
- D
$\sin 2\theta $
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