If 2{sin ^2}x + {sin ^2}2x = 2 for -pi

Question

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

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None of these

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(D) Given equation: $2{\sin ^2}x + {\sin ^2}2x = 2$Using the identity ${\sin ^2}x = \frac{1 - \cos 2x}{2}$,we get:$2(\frac{1 - \cos 2x}{2}) + {\sin ^2}2x = 2$$1 - \cos 2x + {\sin ^2}2x = 2$${ \sin ^2}2x - \cos 2x - 1 = 0$$(1 - {\cos ^2}2x) - \cos 2x - 1 = 0$$- {\cos ^2}2x - \cos 2x = 0$$\cos 2x(\cos 2x + 1) = 0$So,$\cos 2x = 0$ or $\cos 2x = -1$.Case $1$: $\cos 2x = 0$ $\Rightarrow 2x = (2n + 1)\frac{\pi }{2}$ $\Rightarrow x = (2n + 1)\frac{\pi }{4}$.For $-\pi Case $2$: $\cos 2x = -1$ $\Rightarrow 2x = (2n + 1)\pi$ $\Rightarrow x = (2n + 1)\frac{\pi }{2}$.For $-\pi Combining these,$x \in \{ \pm \frac{\pi }{4}, \pm \frac{\pi }{2}, \pm \frac{3\pi }{4} \}$.Since the options provided do not contain the full set,and option $(B)$ is a subset of the solution,the most appropriate choice is $(D)$.

If $2{\sin ^2}x + {\sin ^2}2x = 2$ for $-\pi < x < \pi$,then $x = $

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