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

n π + (- 1)^{n} \frac{π}{2}, n \in Z

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\frac{n π}{2}, n \in Z

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(4 n \pm 1) \frac{π}{2}, n \in Z

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(2 n - 1) π, n \in Z

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Solution

The correct option is C $(4 n \pm 1) \frac{π}{2}, n \in Z$
Given equation is
${sin}^{2} x - cos 2 x = 2 - sin 2 x$
$\Rightarrow 1 - {cos}^{2} x - (2 {cos}^{2} x - 1) = 2 - 2 sin x cos x$
$\Rightarrow cos x (2 sin x - 3 cos x) = 0$
$\Rightarrow cos x = 0$ $(∵ 2 sin x - 3 cos x \neq 0)$
$\Rightarrow x = 4 n \pm 1)$ $\frac{π}{2}$

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