Trigonometrical Equations
medium

यदि $2{\sin ^2}x + {\sin ^2}2x = 2,\, - \pi  < x < \pi ,$ तब $x = $

A

$ \pm \frac{\pi }{6}$

B

$ \pm \frac{\pi }{4}$

C

$\frac{{3\pi }}{2}$

D

इनमें से कोई नहीं

Solution

ज्ञात है, $1 – \cos 2x + 1 – {\cos ^2}2x = 2$

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।

Standard 11
Mathematics

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