For 0 leq x leq pi,if 81^{sin ^2 x}+81^{cos ^ | vedclass

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Question

(A) Given $81^{\sin ^2 x}+81^{\cos ^2 x}=30$. Since $\cos ^2 x = 1 - \sin ^2 x$,we have $81^{\sin ^2 x}+81^{1-\sin ^2 x}=30$. $81^{\sin ^2 x}+\frac{81

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) Given $81^{\sin ^2 x}+81^{\cos ^2 x}=30$. Since $\cos ^2 x = 1 - \sin ^2 x$,we have $81^{\sin ^2 x}+81^{1-\sin ^2 x}=30$. $81^{\sin ^2 x}+\frac{81}{81^{\sin ^2 x}}=30$. Let $t = 81^{\sin ^2 x}$. Then $t + \frac{81}{t} = 30$,which implies $t^2 - 30t + 81 = 0$. Factoring the quadratic equation,we get $(t-3)(t-27) = 0$,so $t = 3$ or $t = 27$. Case $1$: $81^{\sin ^2 x} = 3$ $\Rightarrow 3^{4 \sin ^2 x} = 3^1$ $\Rightarrow 4 \sin ^2 x = 1$ $\Rightarrow \sin ^2 x = \frac{1}{4}$ $\Rightarrow \sin x = \frac{1}{2}$ (for $x \in [0, \pi]$). Thus,$x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$. Case $2$: $81^{\sin ^2 x} = 27$ $\Rightarrow 3^{4 \sin ^2 x} = 3^3$ $\Rightarrow 4 \sin ^2 x = 3$ $\Rightarrow \sin ^2 x = \frac{3}{4}$ $\Rightarrow \sin x = \frac{\sqrt{3}}{2}$. Thus,$x = \frac{\pi}{3}$ or $x = \frac{2\pi}{3}$. Comparing with the given options,$x = \frac{\pi}{6}$ is the correct choice.

For $0 \leq x \leq \pi$,if $81^{\sin ^2 x}+81^{\cos ^2 x}=30$,then $x=$

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