If 3 cos x neq 2 sin x,then the general solut | vedclass

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(A) Given equation: $\sin ^{2} x - \cos 2 x = 2 - \sin 2 x$ Using the identity $\cos 2 x = 1 - 2 \sin ^{2} x$,we get: $\sin ^{2} x - (1 - 2 \sin ^{2}

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$x=n \pi+\frac{\pi}{2}, \quad n \in Z$

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

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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