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

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given equation: $(81)^{\sin ^{2} x} + (81)^{\cos ^{2} x} = 30$.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 $t^{2} - 30t + 81 = 0$.$(t - 3)(t - 27) = 0$,so $t = 3$ or $t = 27$.Case $1$: $(81)^{\sin ^{2} x} = 3 \implies 3^{4 \sin ^{2} x} = 3^{1} \implies 4 \sin ^{2} x = 1 \implies \sin ^{2} x = \frac{1}{4}$.In $[0, \pi]$,$\sin x = \frac{1}{2}$ or $\sin x = -\frac{1}{2}$. Since $\sin x \ge 0$ in $[0, \pi]$,$\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}, \frac{5\pi}{6}$ ($2$ solutions).Case $2$: $(81)^{\sin ^{2} x} = 27 \implies 3^{4 \sin ^{2} x} = 3^{3} \implies 4 \sin ^{2} x = 3 \implies \sin ^{2} x = \frac{3}{4}$.In $[0, \pi]$,$\sin x = \frac{\sqrt{3}}{2}$ or $\sin x = -\frac{\sqrt{3}}{2}$. Since $\sin x \ge 0$ in $[0, \pi]$,$\sin x = \frac{\sqrt{3}}{2}$ gives $x = \frac{\pi}{3}, \frac{2\pi}{3}$ ($2$ solutions).Total number of solutions = $2 + 2 = 4$.

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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