Let S = {theta in [0, 2pi] : 8^{2 sin^2 theta | vedclass

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(C) Let $y = 8^{2 \sin^2 	heta}$. Since $\cos^2 	heta = 1 - \sin^2 	heta$,the equation becomes $y + 8^{2(1 - \sin^2 	heta)} = 16$,which is $y + \f

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$-4$

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(C) Let $y = 8^{2 \sin^2 	heta}$. Since $\cos^2 	heta = 1 - \sin^2 	heta$,the equation becomes $y + 8^{2(1 - \sin^2 	heta)} = 16$,which is $y + \frac{64}{y} = 16$.Multiplying by $y$,we get $y^2 - 16y + 64 = 0$,so $(y - 8)^2 = 0$,which gives $y = 8$.Thus,$8^{2 \sin^2 	heta} = 8^1$,implying $2 \sin^2 	heta = 1$,or $\sin^2 	heta = \frac{1}{2}$.This means $\sin 	heta = \pm \frac{1}{\sqrt{2}}$,so $	heta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$. Thus,$n(S) = 4$.For each $	heta \in S$,$2	heta \in \{\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}\}$.Then $\frac{\pi}{4} + 2	heta \in \{\frac{3\pi}{4}, \frac{7\pi}{4}, \frac{11\pi}{4}, \frac{15\pi}{4}\}$.The expression is $\sum \frac{1}{\cos(\frac{\pi}{4} + 2	heta) \sin(\frac{\pi}{4} + 2	heta)} = \sum \frac{2}{\sin(\frac{\pi}{2} + 4	heta)} = \sum \frac{2}{\cos(4	heta)}$.For all $	heta \in S$,$4	heta$ is an odd multiple of $\pi$,so $\cos(4	heta) = -1$.Thus,the sum is $\sum_{	heta \in S} (-2) = 4 	imes (-2) = -8$.Finally,$n(S) + 	ext{sum} = 4 - 8 = -4$.

Let $S = \{\theta \in [0, 2\pi] : 8^{2 \sin^2 \theta} + 8^{2 \cos^2 \theta} = 16\}$. Then $n(S) + \sum_{\theta \in S} \left(\sec \left(\frac{\pi}{4} + 2\theta\right) \operatorname{cosec} \left(\frac{\pi}{4} + 2\theta\right)\right)$ is equal to.

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