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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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Verified Answer
The correct answer is:
$(4 n \pm 1) \frac{\pi}{2}, n \in Z$
$$
\begin{array}{l}
\sin ^{2} x-\cos 2 x=2-\sin 2 x \\
\Rightarrow 1-\cos ^{2} x-\left(2 \cos ^{2}-1\right)=2-2 \sin x \cos x \\
\Rightarrow-3 \cos ^{2} x+2 \sin x \cos x=0 \\
\Rightarrow \cos x(2 \sin x-3 \cos x)=0 \\
\Rightarrow \cos x=0, \quad(\because 2 \sin x-3 \cos x \neq 0) \\
\Rightarrow x=2 n \pi \pm \frac{\pi}{2} \\
\Rightarrow x=(4 n \pm 1) \frac{\pi}{2}, n \in Z
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow x=2 n \pi \pm \frac{\pi}{2} \\
\Rightarrow x=(4 n \pm 1) \frac{\pi}{2}, n \in Z \\
\end{array}
$$
\begin{array}{l}
\sin ^{2} x-\cos 2 x=2-\sin 2 x \\
\Rightarrow 1-\cos ^{2} x-\left(2 \cos ^{2}-1\right)=2-2 \sin x \cos x \\
\Rightarrow-3 \cos ^{2} x+2 \sin x \cos x=0 \\
\Rightarrow \cos x(2 \sin x-3 \cos x)=0 \\
\Rightarrow \cos x=0, \quad(\because 2 \sin x-3 \cos x \neq 0) \\
\Rightarrow x=2 n \pi \pm \frac{\pi}{2} \\
\Rightarrow x=(4 n \pm 1) \frac{\pi}{2}, n \in Z
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow x=2 n \pi \pm \frac{\pi}{2} \\
\Rightarrow x=(4 n \pm 1) \frac{\pi}{2}, n \in Z \\
\end{array}
$$
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