2{sin ^2}x + {sin ^2}2x = 2,, - pi

Question

(b) We have $1 - \cos 2x + 1 - {\cos ^2}2x = 2$

or $\cos 2x(\cos 2x + 1) = 0$

$	herefore $ $\cos 2x = 0,\, - 1$, 

$	herefore $ $2x = \

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) We have $1 - \cos 2x + 1 - {\cos ^2}2x = 2$

or $\cos 2x(\cos 2x + 1) = 0$

$	herefore $ $\cos 2x = 0,\, - 1$, 

$	herefore $ $2x = \left( {n + \frac{1}{2}} ight){m{ }}\pi \,{m{\, or\,\,}}\,\,(2n + 1){m{ }}\pi $

$ \Rightarrow $ $x = (2n + 1)\frac{\pi }{4}{m{\,\, or\,\, }}(2n + 1)\frac{\pi }{2}$

Now put $n = - 2,\, - 1,\,0,\,1,\,2$

$	herefore $ $x = \frac{{ - 3\pi }}{4},\,\frac{{ - \pi }}{4},\,\frac{\pi }{4},\,\frac{{3\pi }}{4},\,\frac{{5\pi }}{4}$ 

and $\frac{{ - 3\pi }}{2},\,\frac{{ - \pi }}{2},\,\frac{\pi }{2},\,\frac{{3\pi }}{2},\,\frac{{5\pi }}{2}$

Since $ - \pi \le x \le \pi $, 

therefore $x \pm \frac{\pi }{4},\, \pm \frac{\pi }{2},\, \pm \frac{{3\pi }}{4}$ only.

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

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