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(B) Given equation: $\sin^2 	heta + 2 \cos^2 	heta - \sqrt{3} \sin 	heta \cos 	heta = 2$. Using $\sin^2 	heta + \cos^2 	heta = 1$,we can write $

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$3$

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(B) Given equation: $\sin^2 	heta + 2 \cos^2 	heta - \sqrt{3} \sin 	heta \cos 	heta = 2$. Using $\sin^2 	heta + \cos^2 	heta = 1$,we can write $2 = 2(\sin^2 	heta + \cos^2 	heta) = 2 \sin^2 	heta + 2 \cos^2 	heta$. Substituting this into the equation: $\sin^2 	heta + 2 \cos^2 	heta - \sqrt{3} \sin 	heta \cos 	heta = 2 \sin^2 	heta + 2 \cos^2 	heta$. Simplifying,we get: $-\sin^2 	heta - \sqrt{3} \sin 	heta \cos 	heta = 0$. Factor out $-\sin 	heta$: $-\sin 	heta (\sin 	heta + \sqrt{3} \cos 	heta) = 0$. This gives two cases: Case $1$: $\sin 	heta = 0$. In the interval $(-\pi, \pi)$,the solutions are $	heta = 0$. Case $2$: $\sin 	heta + \sqrt{3} \cos 	heta = 0$,which implies $	an 	heta = -\sqrt{3}$. In the interval $(-\pi, \pi)$,$	an 	heta = -\sqrt{3}$ at $	heta = -\frac{\pi}{3}$ and $	heta = \frac{2\pi}{3}$. Thus,the solutions are $\{0, -\frac{\pi}{3}, \frac{2\pi}{3}\}$. The total number of solutions is $3$.

Number of solutions of the equation $\sin^2 \theta + 2 \cos^2 \theta - \sqrt{3} \sin \theta \cos \theta = 2$ lying in the interval $(-\pi, \pi)$ is

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