Question
यदि $2{\sin ^2}x + {\sin ^2}2x = 2,\, - \pi < x < \pi ,$ तब $x = $
या $\cos 2x(\cos 2x + 1) = 0$
$\therefore $ $\cos 2x = 0,\, - 1$,
$\therefore $ $2x = \left( {n + \frac{1}{2}} \right){\rm{ }}\pi \,$ या $(2n + 1){\rm{ }}\pi $
$ \Rightarrow $ $x = (2n + 1)\frac{\pi }{4}{\rm{ }}$ या $(2n + 1)\frac{\pi }{2}$
$n = - 2,\, - 1,\,0,\,1,\,2$ रखने पर,
$\therefore $$x = \frac{{ - 3\pi }}{4},\,\frac{{ - \pi }}{4},\,\frac{\pi }{4},\,\frac{{3\pi }}{4},\,\frac{{5\pi }}{4}$
और $\frac{{ - 3\pi }}{2},\,\frac{{ - \pi }}{2},\,\frac{\pi }{2},\,\frac{{3\pi }}{2},\,\frac{{5\pi }}{2}$
$ - \pi \le x \le \pi $,
$\therefore $ केवल $x = \pm \frac{\pi }{4},\, \pm \frac{\pi }{2},\, \pm \frac{{3\pi }}{4}$ अभीष्ट हल हैं।
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