Let 2sin^2 x + 3sin x - 2 0 and x^2 - x - 2 | vedclass

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(D) Given $2\sin^2 x + 3\sin x - 2 > 0$. Factoring the quadratic: $2\sin^2 x + 4\sin x - \sin x - 2 > 0$. $2\sin x(\sin x + 2) - 1(\sin x + 2) > 0$. $

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

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(D) Given $2\sin^2 x + 3\sin x - 2 > 0$. Factoring the quadratic: $2\sin^2 x + 4\sin x - \sin x - 2 > 0$. $2\sin x(\sin x + 2) - 1(\sin x + 2) > 0$. $(\sin x + 2)(2\sin x - 1) > 0$. Since $\sin x + 2 > 0$ for all real $x$,we must have $2\sin x - 1 > 0$,which implies $\sin x > 1/2$. For $x$ in a reasonable range,$\sin x > 1/2$ implies $x > \pi/6$. Given $x^2 - x - 2 Factoring the quadratic: $(x - 2)(x + 1) This inequality holds for $x \in (-1, 2)$. Combining the two conditions: $x > \pi/6$ and $-1 Since $\pi/6 \approx 0.523$,the intersection is $x \in (\pi/6, 2)$.

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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