The sum of all values of theta in [0, 2pi] sa | vedclass

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(D) Given equations are $2 \sin^2 	heta = \cos 2	heta$ and $2 \cos^2 	heta = 3 \sin 	heta$. From the first equation: $2 \sin^2 	heta = 1 - 2 \sin

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$\pi$

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(D) Given equations are $2 \sin^2 	heta = \cos 2	heta$ and $2 \cos^2 	heta = 3 \sin 	heta$. From the first equation: $2 \sin^2 	heta = 1 - 2 \sin^2 	heta \implies 4 \sin^2 	heta = 1 \implies \sin^2 	heta = \frac{1}{4} \implies \sin 	heta = \pm \frac{1}{2}$. From the second equation: $2(1 - \sin^2 	heta) = 3 \sin 	heta \implies 2 - 2 \sin^2 	heta = 3 \sin 	heta \implies 2 \sin^2 	heta + 3 \sin 	heta - 2 = 0$. Factoring the quadratic: $(2 \sin 	heta - 1)(\sin 	heta + 2) = 0$. Since $\sin 	heta$ cannot be $-2$,we have $\sin 	heta = \frac{1}{2}$. Comparing both results,the common solution is $\sin 	heta = \frac{1}{2}$. For $	heta \in [0, 2\pi]$,the values are $	heta = \frac{\pi}{6}$ and $	heta = \frac{5\pi}{6}$. The sum of these values is $\frac{\pi}{6} + \frac{5\pi}{6} = \pi$.

The sum of all values of $\theta \in [0, 2\pi]$ satisfying $2 \sin^2 \theta = \cos 2\theta$ and $2 \cos^2 \theta = 3 \sin \theta$ is

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