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

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(D) $2 \sin^2 x + 3 \sin x - 2 > 0$ . . . . . . $(i)$ Let $y = \sin x$. Then the inequality becomes $2 y^2 + 3 y - 2 > 0$. Factorizing,we get $(2 y -

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

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(D) $2 \sin^2 x + 3 \sin x - 2 > 0$ . . . . . . $(i)$ Let $y = \sin x$. Then the inequality becomes $2 y^2 + 3 y - 2 > 0$. Factorizing,we get $(2 y - 1)(y + 2) > 0$. Since $\sin x$ is always $\geq -1$,the term $(y + 2)$ is always positive. Therefore,$2 y - 1 > 0$,which implies $\sin x > \frac{1}{2}$. For $x$ in a standard range,this gives $x \in \left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$. Given $x^2 - x - 2 This implies $x \in (-1, 2)$. Taking the intersection of $x \in \left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$ and $x \in (-1, 2)$,since $\frac{\pi}{6} \approx 0.52$ and $\frac{5 \pi}{6} \approx 2.61$,the intersection is $\left(\frac{\pi}{6}, 2\right)$.

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

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