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

\[\begin {array} {1 1} (1)\;n \pi +(-1)^n \frac{\pi}{2}, n \in Z & \quad (2)\;\frac{n\pi}{2}, n \in Z \\ (3)\;(4 n \pm 1) \frac{\pi}{2}, n \in Z & \quad (4)\;(2n-1) \pi, n \in Z \end {array}\]
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$ (3)\;(4 n \pm 1) \frac{\pi}{2}, n \in Z$
answered Nov 7, 2013 by pady_1

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