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

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

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$(4 n \pm 1) \frac{\pi}{2}, n \in \mathbb{Z}$

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(C) Given equation: $\sin^2 x - \cos 2x = 2 - \sin 2x$ Using identities $\cos 2x = 1 - 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$: $\sin^2 x - (1 - 2\sin^2 x) = 2 - 2\sin x \cos x$ $3\sin^2 x - 1 = 2 - 2\sin x \cos x$ Alternatively,using $\cos 2x = 2\cos^2 x - 1$: $\sin^2 x - (2\cos^2 x - 1) = 2 - 2\sin x \cos x$ $(1 - \cos^2 x) - 2\cos^2 x + 1 = 2 - 2\sin x \cos x$ $2 - 3\cos^2 x = 2 - 2\sin x \cos x$ $-3\cos^2 x + 2\sin x \cos x = 0$ $\cos x (2\sin x - 3\cos x) = 0$ Since $3\cos x 
eq 2\sin x$,we must have $\cos x = 0$. The general solution for $\cos x = 0$ is $x = (2n + 1) \frac{\pi}{2}$,which can be written as $x = (4n \pm 1) \frac{\pi}{2}$ for $n \in \mathbb{Z}$.

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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