If 0 le x le pi and 81^{sin^2 x} + 81^{cos^2 | vedclass

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Question

(A) Given the equation: $81^{\sin^2 x} + 81^{\cos^2 x} = 30$. Let $u = 81^{\sin^2 x}$. Since $\cos^2 x = 1 - \sin^2 x$,the equation becomes $u + 81^{1

Vedclass · Accepted Answer

$\pi /6$

Vedclass · Answer

(A) Given the equation: $81^{\sin^2 x} + 81^{\cos^2 x} = 30$. Let $u = 81^{\sin^2 x}$. Since $\cos^2 x = 1 - \sin^2 x$,the equation becomes $u + 81^{1 - \sin^2 x} = 30$. This simplifies to $u + \frac{81}{u} = 30$. Multiplying by $u$,we get $u^2 - 30u + 81 = 0$. Factoring the quadratic: $(u - 27)(u - 3) = 0$. So,$u = 27$ or $u = 3$. Case $1$: $81^{\sin^2 x} = 27 \implies (3^4)^{\sin^2 x} = 3^3 \implies 4 \sin^2 x = 3 \implies \sin^2 x = 3/4 \implies \sin x = \pm \sqrt{3}/2$. For $0 \le x \le \pi$,$x = \pi/3$ or $x = 2\pi/3$. Case $2$: $81^{\sin^2 x} = 3 \implies (3^4)^{\sin^2 x} = 3^1 \implies 4 \sin^2 x = 1 \implies \sin^2 x = 1/4 \implies \sin x = \pm 1/2$. For $0 \le x \le \pi$,$x = \pi/6$ or $x = 5\pi/6$. Checking the options,$\pi/6$ is the correct value.

If $0 \le x \le \pi$ and $81^{\sin^2 x} + 81^{\cos^2 x} = 30$,then $x =$

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