Let S = {theta in (-2pi, 2pi) : costheta + 1 | vedclass

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(B) Given equation: $\cos	heta - \sqrt{3}\sin	heta = -1$.Divide by $2$: $\frac{1}{2}\cos	heta - \frac{\sqrt{3}}{2}\sin	heta = -\frac{1}{2}$.This c

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$-\frac{4\pi}{3}$

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(B) Given equation: $\cos	heta - \sqrt{3}\sin	heta = -1$.Divide by $2$: $\frac{1}{2}\cos	heta - \frac{\sqrt{3}}{2}\sin	heta = -\frac{1}{2}$.This can be written as $\cos(	heta + \frac{\pi}{3}) = -\frac{1}{2}$.Let $\alpha = 	heta + \frac{\pi}{3}$. Since $	heta \in (-2\pi, 2\pi)$,$\alpha \in (-2\pi + \frac{\pi}{3}, 2\pi + \frac{\pi}{3}) = (-\frac{5\pi}{3}, \frac{7\pi}{3})$.For $\cos\alpha = -\frac{1}{2}$,the solutions are $\alpha = \frac{2\pi}{3} + 2n\pi$ or $\alpha = \frac{4\pi}{3} + 2n\pi$.For $n=0$: $\alpha = \frac{2\pi}{3}, \frac{4\pi}{3}$.For $n=1$: $\alpha = \frac{8\pi}{3}$ (out of range),$\alpha = \frac{10\pi}{3}$ (out of range).For $n=-1$: $\alpha = \frac{2\pi}{3} - 2\pi = -\frac{4\pi}{3}$,$\alpha = \frac{4\pi}{3} - 2\pi = -\frac{2\pi}{3}$.Thus,$	heta = \alpha - \frac{\pi}{3}$ gives $	heta = \frac{\pi}{3}, \pi, -\frac{5\pi}{3}, -\pi$.The sum of these values is $\frac{\pi}{3} + \pi - \frac{5\pi}{3} - \pi = -\frac{4\pi}{3}$.

Let $S = \{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3} \sin\theta\}$. Then $\sum_{\theta \in S} \theta$ is equal to:

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