Let $2 \sin ^2 x+3 \sin x-2\gt0$ and $x^2-x-2 \lt 0$ ( $x$ is measured in radians). Then $x$ lies in the interval

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Aicharya · Accepted Answer

$\begin{aligned} 	herefore \quad & 2 \sin ^2 x+3 \sin x-2\gt0 \ & (2 \sin x-1)(\sin x+2)\gt0 \ & \Rightarrow 2 \sin x-1\gt0 \ldots[\because \sin x+2\gt0 	ext { and } x \in \mathrm{R}] \ & \Rightarrow \sin x\gt\frac{1}{2} \ & \Rightarrow x \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}ight) \end{aligned}$ $\begin{aligned} & 	ext { Also, } x^2-x-2 \lt 0 \ & \Rightarrow(x-2)(x+1) \lt 0 \ & \Rightarrow-1 \lt x \lt 2 \end{aligned}$ Since $2 \lt \frac{5 \pi}{6}$ $	herefore \quad x$ must lie in $\left(\frac{\pi}{6}, 2ight)$

Let $2 \sin ^2 x+3 \sin x-2\gt0$ and $x^2-x-2 \lt 0$ ( $x$ is measured in radians). Then $x$ lies in the…

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