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

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(A) Given equation: $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x) + 2\cos x = 0$Substitute $\cos 2x = 1 - 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$:$2(\

Vedclass · Accepted Answer

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

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(A) Given equation: $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x) + 2\cos x = 0$Substitute $\cos 2x = 1 - 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$:$2(\sin x - (1 - 2\sin^2 x)) - 2\sin x \cos x(1 + 2\sin x) + 2\cos x = 0$$2\sin x - 2 + 4\sin^2 x - 2\sin x \cos x - 4\sin^2 x \cos x + 2\cos x = 0$Group terms:$(4\sin^2 x + 2\sin x - 2) - \cos x(2\sin x + 4\sin^2 x - 2) = 0$$(4\sin^2 x + 2\sin x - 2)(1 - \cos x) = 0$$2(2\sin^2 x + \sin x - 1)(1 - \cos x) = 0$$2(2\sin x - 1)(\sin x + 1)(1 - \cos x) = 0$This gives $\sin x = \frac{1}{2}$,$\sin x = -1$,or $\cos x = 1$.For $\sin x = \frac{1}{2}$,$x = n\pi + (-1)^n \frac{\pi}{6}$,which simplifies to $x = \frac{\pi}{6}(4n + 1)$ or $x = \frac{\pi}{6}(4n + 5)$.For $\sin x = -1$,$x = 2n\pi - \frac{\pi}{2} = \frac{\pi}{2}(4n - 1)$.For $\cos x = 1$,$x = 2n\pi = \frac{\pi}{2}(4n)$.The solution set matches option $A$.

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

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