If for certain x,3 cos x neq 2 sin x,then the | vedclass

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(A) Given equation: $\sin ^2 x - \cos 2 x = 2 - \sin 2 x$ Using identities $\cos 2 x = 1 - 2 \sin ^2 x$ and $\sin 2 x = 2 \sin x \cos x$: $\sin ^2 x -

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$(2 n + 1) \frac{\pi}{2}, n \in Z$

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(A) Given equation: $\sin ^2 x - \cos 2 x = 2 - \sin 2 x$ Using identities $\cos 2 x = 1 - 2 \sin ^2 x$ and $\sin 2 x = 2 \sin x \cos x$: $\sin ^2 x - (1 - 2 \sin ^2 x) = 2 - 2 \sin x \cos x$ $3 \sin ^2 x - 1 = 2 - 2 \sin x \cos x$ $3 \sin ^2 x + 2 \sin x \cos x - 3 = 0$ Since $\sin ^2 x + \cos ^2 x = 1$,we have $3 \sin ^2 x + 2 \sin x \cos x - 3(\sin ^2 x + \cos ^2 x) = 0$ $3 \sin ^2 x + 2 \sin x \cos x - 3 \sin ^2 x - 3 \cos ^2 x = 0$ $2 \sin x \cos x - 3 \cos ^2 x = 0$ $\cos x (2 \sin x - 3 \cos x) = 0$ This implies $\cos x = 0$ or $2 \sin x - 3 \cos x = 0$. Given $3 \cos x 
eq 2 \sin x$,we must have $\cos x = 0$. The general solution for $\cos x = 0$ is $x = (2 n + 1) \frac{\pi}{2}, n \in Z$.

If for certain $x$,$3 \cos x \neq 2 \sin x$,then the general solution of $\sin ^2 x - \cos 2 x = 2 - \sin 2 x$ is

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