MCQ
જો $4{\sin ^4}x + {\cos ^4}x = 1,$ તો $x =$
- ✓$n\pi $
- B$n\pi \pm {\sin ^{ - 1}}\frac{2}{5}$
- C$n\pi + \frac{\pi }{6}$
- Dએકપણ નહિ.
$4{\sin ^4}x = 1 - {\cos ^4}x = (1 - {\cos ^2}x)\,(1 + {\cos ^2}x)$
$ \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$ or $\sin x = \pm \sqrt {2/5} $.
Hence $x = n\pi $ is the required answer.
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