The number of solutions of 16^{sin ^2 x} + 16 | vedclass

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Question

(A) Let $y = 16^{\sin ^2 x}$. Since $\cos ^2 x = 1 - \sin ^2 x$,we have $16^{\cos ^2 x} = 16^{1 - \sin ^2 x} = \frac{16}{16^{\sin ^2 x}} = \frac{16}{y

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let $y = 16^{\sin ^2 x}$. Since $\cos ^2 x = 1 - \sin ^2 x$,we have $16^{\cos ^2 x} = 16^{1 - \sin ^2 x} = \frac{16}{16^{\sin ^2 x}} = \frac{16}{y}$.Substituting this into the equation: $y + \frac{16}{y} = 10$.Multiplying by $y$: $y^2 - 10y + 16 = 0$.Factoring the quadratic: $(y - 8)(y - 2) = 0$,so $y = 8$ or $y = 2$.Case $1$: $16^{\sin ^2 x} = 8 \implies (2^4)^{\sin ^2 x} = 2^3 \implies 4 \sin ^2 x = 3 \implies \sin ^2 x = \frac{3}{4} \implies \sin x = \pm \frac{\sqrt{3}}{2}$.In $0 \leqslant x \leqslant 2\pi$,$\sin x = \frac{\sqrt{3}}{2}$ at $x = \frac{\pi}{3}, \frac{2\pi}{3}$ and $\sin x = -\frac{\sqrt{3}}{2}$ at $x = \frac{4\pi}{3}, \frac{5\pi}{3}$. ($4$ solutions).Case $2$: $16^{\sin ^2 x} = 2 \implies (2^4)^{\sin ^2 x} = 2^1 \implies 4 \sin ^2 x = 1 \implies \sin ^2 x = \frac{1}{4} \implies \sin x = \pm \frac{1}{2}$.In $0 \leqslant x \leqslant 2\pi$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}, \frac{5\pi}{6}$ and $\sin x = -\frac{1}{2}$ at $x = \frac{7\pi}{6}, \frac{11\pi}{6}$. ($4$ solutions).Total number of solutions = $4 + 4 = 8$.

The number of solutions of $16^{\sin ^2 x} + 16^{\cos ^2 x} = 10$ in the interval $0 \leqslant x \leqslant 2\pi$ is:

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