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(A) Given equations are $2 \sin^{2} 	heta - \cos 2	heta = 0$ and $2 \cos^{2} 	heta + 3 \sin 	heta = 0$.For the first equation:$2 \sin^{2} 	heta -

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$3$

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(A) Given equations are $2 \sin^{2} 	heta - \cos 2	heta = 0$ and $2 \cos^{2} 	heta + 3 \sin 	heta = 0$.For the first equation:$2 \sin^{2} 	heta - (1 - 2 \sin^{2} 	heta) = 0$$4 \sin^{2} 	heta = 1$$\sin^{2} 	heta = \frac{1}{4} \implies \sin 	heta = \pm \frac{1}{2}$.In $[0, 2\pi]$,$	heta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.For the second equation:$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 $\sin 	heta = -\frac{1}{2}$.In $[0, 2\pi]$,$	heta = \frac{7\pi}{6}, \frac{11\pi}{6}$.The common solutions are $	heta = \frac{7\pi}{6}, \frac{11\pi}{6}$.Sum of solutions $= \frac{7\pi}{6} + \frac{11\pi}{6} = \frac{18\pi}{6} = 3\pi$.Given sum $= k\pi$,so $k = 3$.

If the sum of solutions of the system of equations $2 \sin^{2} \theta - \cos 2\theta = 0$ and $2 \cos^{2} \theta + 3 \sin \theta = 0$ in the interval $[0, 2\pi]$ is $k\pi$,then $k$ is equal to.

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