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(D) Given equation: $2\sin^2 	heta - 3\sin 	heta - 2 = 0$Factorizing the quadratic equation: $(2\sin 	heta + 1)(\sin 	heta - 2) = 0$Since $\sin

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$n\pi + (-1)^n \frac{7\pi}{6}$

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(D) Given equation: $2\sin^2 	heta - 3\sin 	heta - 2 = 0$Factorizing the quadratic equation: $(2\sin 	heta + 1)(\sin 	heta - 2) = 0$Since $\sin 	heta$ cannot be $2$,we have $2\sin 	heta + 1 = 0$,which implies $\sin 	heta = -\frac{1}{2}$.We know that $\sin 	heta = -\frac{1}{2} = \sin(-\frac{\pi}{6})$.The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$.Thus,$	heta = n\pi + (-1)^n (-\frac{\pi}{6})$.Using the property $(-1)^n (-1) = (-1)^{n+1}$,we can write $	heta = n\pi + (-1)^{n+1} \frac{\pi}{6}$.Alternatively,since $-\frac{\pi}{6}$ is equivalent to $\frac{11\pi}{6}$ or $\frac{7\pi}{6}$ in different contexts,checking the options,the form $n\pi + (-1)^n \frac{7\pi}{6}$ is mathematically consistent with the general solution set.

The general value of $\theta$ satisfying the equation $2\sin^2 \theta - 3\sin \theta - 2 = 0$ is

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