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(D) Given $2 \sin^2 x + 3 \sin x - 2 > 0$. Factoring the quadratic expression,we get $(2 \sin x - 1)(\sin x + 2) > 0$. Since $\sin x + 2 > 0$ for all

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$(\frac{\pi}{6}, 2)$

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(D) Given $2 \sin^2 x + 3 \sin x - 2 > 0$. Factoring the quadratic expression,we get $(2 \sin x - 1)(\sin x + 2) > 0$. Since $\sin x + 2 > 0$ for all $x \in \mathbb{R}$,we must have $2 \sin x - 1 > 0$,which implies $\sin x > \frac{1}{2}$. This gives $x \in (\frac{\pi}{6}, \frac{5 \pi}{6})$. Also,given $x^2 - x - 2 Factoring gives $(x - 2)(x + 1) To find the intersection,we note that $\frac{\pi}{6} \approx 0.52$ and $\frac{5 \pi}{6} \approx 2.61$. Since $2 < \frac{5 \pi}{6}$,the intersection of $(\frac{\pi}{6}, \frac{5 \pi}{6})$ and $(-1, 2)$ is $(\frac{\pi}{6}, 2)$.

Let $2 \sin^2 x + 3 \sin x - 2 > 0$ and $x^2 - x - 2 < 0$ ($x$ is measured in radians). Then $x$ lies in the interval

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