If 2(sin x - cos 2x) - sin 2x(1 + 2sin x)2cos | vedclass

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Question

(a) $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x) + 2\cos x = 0$
==> $2\sin x - 2 + 4{\sin ^2}x - 2\sin x\cos x - 4{\sin ^2}x\cos x$
$ + 2\cos x =

Vedclass · Accepted Answer

$x = \frac{\pi }{6}(4n + 1)$or $x = \frac{\pi }{2}(4n - 1)$

Vedclass · Answer

(a) $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x) + 2\cos x = 0$
==> $2\sin x - 2 + 4{\sin ^2}x - 2\sin x\cos x - 4{\sin ^2}x\cos x$
$ + 2\cos x = 0$
==>$4{\sin ^2}x + 2\sin x - 2 - \cos x[4{\sin ^2}x + 2\sin x - 2]$= 0
==> $(1 - \cos x)\,(\sin x + 1)\,(4\sin x - 2) = 0$
Hence $\sin x = - 1$ or $\cos x = 1$ or $\sin x = \frac{1}{2}$
==> $x = (4n - 1)\,\frac{\pi }{2}$ and $x = (4n + 1)\frac{\pi }{6}$.

If $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x)2\cos x = 0$ then

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