Question
यदि $4{\sin ^4}x + {\cos ^4}x = 1,$ तब $x = $
$4{\sin ^4}x = 1 - {\cos ^4}x = (1 - {\cos ^2}x)\,(1 + {\cos ^2}x)$
$ \Rightarrow $ ${\sin ^2}x[4{\sin ^2}x - 1 - (1 - {\sin ^2}x)] = 0$
$ \Rightarrow $ ${\sin ^2}x[5{\sin ^2}x - 2] = 0$
$ \Rightarrow $$\sin x = 0$ या $\sin x = \pm \sqrt {\frac{2}{5}} $
अत: $x = n\pi $ अभीष्ट हल है।
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