The number of solutions in [0, 2pi] of the eq | vedclass

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(D) Given equation: $16^{\sin^2 x} + 16^{\cos^2 x} = 10$ Since $\cos^2 x = 1 - \sin^2 x$,we have: $16^{\sin^2 x} + 16^{1 - \sin^2 x} = 10$ $16^{\sin^2

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$8$

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(D) Given equation: $16^{\sin^2 x} + 16^{\cos^2 x} = 10$ Since $\cos^2 x = 1 - \sin^2 x$,we have: $16^{\sin^2 x} + 16^{1 - \sin^2 x} = 10$ $16^{\sin^2 x} + \frac{16}{16^{\sin^2 x}} = 10$ Let $t = 16^{\sin^2 x}$. Then $t + \frac{16}{t} = 10$,which implies $t^2 - 10t + 16 = 0$. Solving for $t$: $(t - 8)(t - 2) = 0$,so $t = 8$ or $t = 2$. Case $1$: $16^{\sin^2 x} = 2$ $\Rightarrow 2^{4\sin^2 x} = 2^1$ $\Rightarrow 4\sin^2 x = 1$ $\Rightarrow \sin^2 x = \frac{1}{4}$ $\Rightarrow \sin x = \pm \frac{1}{2}$. In $[0, 2\pi]$,$\sin x = \pm \frac{1}{2}$ gives $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ ($4$ solutions). Case $2$: $16^{\sin^2 x} = 8$ $\Rightarrow 2^{4\sin^2 x} = 2^3$ $\Rightarrow 4\sin^2 x = 3$ $\Rightarrow \sin^2 x = \frac{3}{4}$ $\Rightarrow \sin x = \pm \frac{\sqrt{3}}{2}$. In $[0, 2\pi]$,$\sin x = \pm \frac{\sqrt{3}}{2}$ gives $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$ ($4$ solutions). Total number of solutions = $4 + 4 = 8$.

The number of solutions in $[0, 2\pi]$ of the equation $16^{\sin^2 x} + 16^{\cos^2 x} = 10$ is

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