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If $3 \cos x \neq 2 \sin x$, then the general solution of $\sin ^{2} x-\cos 2 x=2-\sin 2 x$ is
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The correct answer is:
$x=n \pi+\frac{\pi}{2}, \quad n \in Z$
Given $\sin ^{2} x-\cos 2 x=2-\sin 2 x$
$\therefore \sin ^{2} x-1+2 \sin ^{2} x=2-\sin 2 x \quad \ldots\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]$
$3 \sin ^{2} x \quad=3-\sin 2 x \Rightarrow \sin 2 x=3\left(1-\sin ^{2} x\right)$
$2 \sin x \cos x=3 \cos ^{2} x \Rightarrow \cos x(3 \cos x-2 \sin x)=0$
$\therefore \cos x=0$ or $3 \cos x=2 \sin x$
But $3 \cos x \neq 2 \sin x$ as per condition given
$\therefore \cos \mathrm{x}=0 \Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$
$\therefore \sin ^{2} x-1+2 \sin ^{2} x=2-\sin 2 x \quad \ldots\left[\because \cos 2 \theta=1-2 \sin ^{2} \theta\right]$
$3 \sin ^{2} x \quad=3-\sin 2 x \Rightarrow \sin 2 x=3\left(1-\sin ^{2} x\right)$
$2 \sin x \cos x=3 \cos ^{2} x \Rightarrow \cos x(3 \cos x-2 \sin x)=0$
$\therefore \cos x=0$ or $3 \cos x=2 \sin x$
But $3 \cos x \neq 2 \sin x$ as per condition given
$\therefore \cos \mathrm{x}=0 \Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$
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