The sum of the solutions of cos x sqrt{16 sin | vedclass

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(D) Given the equation $\cos x \sqrt{16 \sin ^2 x} = 1$. This simplifies to $\cos x \cdot 4 |\sin x| = 1$,or $2 \sin(2x) = 1$ (when $\sin x > 0$) and

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$4 \pi$

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(D) Given the equation $\cos x \sqrt{16 \sin ^2 x} = 1$. This simplifies to $\cos x \cdot 4 |\sin x| = 1$,or $2 \sin(2x) = 1$ (when $\sin x > 0$) and $-2 \sin(2x) = 1$ (when $\sin x Case $1$: $\sin x > 0$ $(x \in (0, \pi))$. Then $2 \sin(2x) = 1 \implies \sin(2x) = \frac{1}{2}$. $2x = \frac{\pi}{6}, \frac{5 \pi}{6} \implies x = \frac{\pi}{12}, \frac{5 \pi}{12}$. Both are in $(0, \pi)$. Case $2$: $\sin x $2x = \frac{7 \pi}{6}, \frac{11 \pi}{6} \implies x = \frac{7 \pi}{12}, \frac{11 \pi}{12}$. However,these values are in $(0, \pi)$,where $\sin x > 0$,so they are rejected. Wait,checking $2x$ range for $x \in (\pi, 2 \pi)$ is $2x \in (2 \pi, 4 \pi)$. $2x = 2 \pi + \frac{7 \pi}{6} = \frac{19 \pi}{6} \implies x = \frac{19 \pi}{12}$ and $2x = 2 \pi + \frac{11 \pi}{6} = \frac{23 \pi}{6} \implies x = \frac{23 \pi}{12}$. Sum of solutions $= \frac{\pi}{12} + \frac{5 \pi}{12} + \frac{19 \pi}{12} + \frac{23 \pi}{12} = \frac{48 \pi}{12} = 4 \pi$.

The sum of the solutions of $\cos x \sqrt{16 \sin ^2 x} = 1$ in $(0, 2 \pi)$ is

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