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(B) Step $1$: Solve the first equation $2 \sin^2 	heta - 2 \cos^2 	heta + 1 = 0$.Using $\cos^2 	heta = 1 - \sin^2 	heta$, we get $2 \sin^2 	heta

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$2$

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(B) Step $1$: Solve the first equation $2 \sin^2 	heta - 2 \cos^2 	heta + 1 = 0$.Using $\cos^2 	heta = 1 - \sin^2 	heta$, we get $2 \sin^2 	heta - 2(1 - \sin^2 	heta) + 1 = 0$.$2 \sin^2 	heta - 2 + 2 \sin^2 	heta + 1 = 0 \implies 4 \sin^2 	heta = 1 \implies \sin^2 	heta = \frac{1}{4} \implies \sin 	heta = \pm \frac{1}{2}$.Step $2$: Solve the second equation $2 \sin^2 	heta + 3 \sin 	heta - 2 = 0$.Let $x = \sin 	heta$. Then $2x^2 + 3x - 2 = 0 \implies (2x - 1)(x + 2) = 0$.So, $\sin 	heta = \frac{1}{2}$ or $\sin 	heta = -2$. Since $-1 \le \sin 	heta \le 1$, we have $\sin 	heta = \frac{1}{2}$.Step $3$: Find common solutions.The common condition is $\sin 	heta = \frac{1}{2}$.In the interval $[0, 2\pi]$, $\sin 	heta = \frac{1}{2}$ at $	heta = \frac{\pi}{6}$ and $	heta = \frac{5\pi}{6}$.There are $2$ common solutions.

The number of common solutions of the pair of equations $2 \sin^2 \theta - 2 \cos^2 \theta + 1 = 0$ and $2 \sin^2 \theta + 3 \sin \theta - 2 = 0$ in the interval $[0, 2\pi]$ is:

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