If 4{sin ^4}x + {cos ^4}x = 1, then x = | vedclass

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Question

(a) The given equation can be put in the form

$4{\sin ^4}x = 1 - {\cos ^4}x = (1 - {\cos ^2}x)\,(1 + {\cos ^2}x)$

$ \Rightarrow $ ${\sin ^2}x[4{

Vedclass · Accepted Answer

$n\pi $

Vedclass · Answer

(a) The given equation can be put in the form

$4{\sin ^4}x = 1 - {\cos ^4}x = (1 - {\cos ^2}x)\,(1 + {\cos ^2}x)$

$ \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.

If $4{\sin ^4}x + {\cos ^4}x = 1,$ then $x =$

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