If 0 leqslant x leqslant pi and 81^{sin ^2 x} | vedclass

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Question

(A) Let $u = 81^{\sin^2 x}$. Since $\cos^2 x = 1 - \sin^2 x$,we have $81^{\cos^2 x} = 81^{1 - \sin^2 x} = \frac{81}{81^{\sin^2 x}} = \frac{81}{u}$.Sub

Vedclass · Accepted Answer

$\frac{\pi}{6}, \frac{5\pi}{6}$

Vedclass · Answer

(A) Let $u = 81^{\sin^2 x}$. Since $\cos^2 x = 1 - \sin^2 x$,we have $81^{\cos^2 x} = 81^{1 - \sin^2 x} = \frac{81}{81^{\sin^2 x}} = \frac{81}{u}$.Substituting into the equation: $u + \frac{81}{u} = 30$.Multiplying by $u$: $u^2 - 30u + 81 = 0$.Factoring the quadratic: $(u - 27)(u - 3) = 0$.So,$u = 27$ or $u = 3$.Case $1$: $81^{\sin^2 x} = 27 \implies (3^4)^{\sin^2 x} = 3^3 \implies 4\sin^2 x = 3 \implies \sin^2 x = \frac{3}{4} \implies \sin x = \pm \frac{\sqrt{3}}{2}$.For $0 \leqslant x \leqslant \pi$,$x = \frac{\pi}{3}$ or $x = \frac{2\pi}{3}$.Case $2$: $81^{\sin^2 x} = 3 \implies (3^4)^{\sin^2 x} = 3^1 \implies 4\sin^2 x = 1 \implies \sin^2 x = \frac{1}{4} \implies \sin x = \pm \frac{1}{2}$.For $0 \leqslant x \leqslant \pi$,$x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$.Comparing with options,the set $\{\frac{\pi}{6}, \frac{5\pi}{6}\}$ is a valid solution set.

If $0 \leqslant x \leqslant \pi$ and $81^{\sin ^2 x} + 81^{\cos ^2 x} = 30$,then $x$ takes the value:

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