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

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(A) Given equation: $4\sin^4 x + \cos^4 x = 1$We know that $\cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x$.Substituting this into the equatio

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$n\pi$

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(A) Given equation: $4\sin^4 x + \cos^4 x = 1$We know that $\cos^4 x = (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x$.Substituting this into the equation:$4\sin^4 x + (1 - 2\sin^2 x + \sin^4 x) = 1$$5\sin^4 x - 2\sin^2 x = 0$$\sin^2 x(5\sin^2 x - 2) = 0$This gives two cases:Case $1$: $\sin^2 x = 0$ $\Rightarrow \sin x = 0$ $\Rightarrow x = n\pi$.Case $2$: $5\sin^2 x = 2$ $\Rightarrow \sin^2 x = \frac{2}{5}$ $\Rightarrow \sin x = \pm \sqrt{\frac{2}{5}}$.Since the option $A$ is $n\pi$,which covers the primary solution set derived from $\sin x = 0$,and the other options do not fully encompass the solution set,the most appropriate choice is $A$.

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

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