Let S = { theta in [ - 2pi , 2pi ] : 2cos^2 t | vedclass

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(B) Given equation: $2\cos^2 	heta + 3\sin 	heta = 0$Using $\cos^2 	heta = 1 - \sin^2 	heta$,we get:$2(1 - \sin^2 	heta) + 3\sin 	heta = 0$$2 -

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$2\pi$

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(B) Given equation: $2\cos^2 	heta + 3\sin 	heta = 0$Using $\cos^2 	heta = 1 - \sin^2 	heta$,we get:$2(1 - \sin^2 	heta) + 3\sin 	heta = 0$$2 - 2\sin^2 	heta + 3\sin 	heta = 0$$2\sin^2 	heta - 3\sin 	heta - 2 = 0$$(2\sin 	heta + 1)(\sin 	heta - 2) = 0$Since $\sin 	heta = 2$ is impossible,we have $\sin 	heta = -\frac{1}{2}$.For $	heta \in [-2\pi, 2\pi]$,the solutions for $\sin 	heta = -\frac{1}{2}$ are:$	heta = -\frac{\pi}{6}, -\frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.Sum of elements = $(-\frac{\pi}{6}) + (-\frac{5\pi}{6}) + (\frac{7\pi}{6}) + (\frac{11\pi}{6}) = \frac{-6\pi + 18\pi}{6} = \frac{12\pi}{6} = 2\pi$.

Let $S = \{ \theta \in [ - 2\pi , 2\pi ] : 2\cos^2 \theta + 3\sin \theta = 0 \}$. Then the sum of the elements of $S$ is

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