The number of roots of the equation (81)^{sin | vedclass

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Question

(A) Given equation: $(81)^{\sin ^2 x} + (81)^{\cos ^2 x} = 30 \dots (i)$ Let $y = 81^{\sin ^2 x}$. Then $81^{\cos ^2 x} = 81^{(1 - \sin ^2 x)} = \frac

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given equation: $(81)^{\sin ^2 x} + (81)^{\cos ^2 x} = 30 \dots (i)$ Let $y = 81^{\sin ^2 x}$. Then $81^{\cos ^2 x} = 81^{(1 - \sin ^2 x)} = \frac{81}{81^{\sin ^2 x}} = \frac{81}{y}$. Substituting into $(i)$,we get $y + \frac{81}{y} = 30$. $y^2 - 30y + 81 = 0$. $(y - 27)(y - 3) = 0$. So,$y = 27$ or $y = 3$. Case $1$: $81^{\sin ^2 x} = 27 \implies 3^{4 \sin ^2 x} = 3^3 \implies 4 \sin ^2 x = 3 \implies \sin ^2 x = \frac{3}{4} \implies \sin x = \pm \frac{\sqrt{3}}{2}$. Since $x \in [0, \pi]$,$\sin x$ must be positive,so $\sin x = \frac{\sqrt{3}}{2} \implies x = \frac{\pi}{3}, \frac{2\pi}{3}$. Case $2$: $81^{\sin ^2 x} = 3 \implies 3^{4 \sin ^2 x} = 3^1 \implies 4 \sin ^2 x = 1 \implies \sin ^2 x = \frac{1}{4} \implies \sin x = \pm \frac{1}{2}$. Since $x \in [0, \pi]$,$\sin x = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}$. The roots are $\frac{\pi}{3}, \frac{2\pi}{3}, \frac{\pi}{6}, \frac{5\pi}{6}$. Thus,the total number of roots is $4$.

The number of roots of the equation $(81)^{\sin ^2 x} + (81)^{\cos ^2 x} = 30$ in the interval $[0, \pi]$ is equal to

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