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

\(2 \sin ^2 x+3 \sin x-2\gt0...(i)\)
Let \(y=\sin x\)
\(\therefore\) Equation (i) becomes
\(2 y^2+3 y-2\gt0 \)
\( \therefore 2 y^2+4 y-y-2\gt0 \)
\( \therefore 2 y(y+2)-1(y+2)\gt0 \)
\( \therefore (2 y-1)(y+2)\gt0 \)
\( \therefore (2 y-1)\gt0 \text { and }(y+2)\gt0 \text { OR } \)
\( (2 y-1) \lt 0 \text { and }(y+2) \lt 0\)
\(\therefore y\gt\frac{1}{2} \text { and } y\gt-2 \)
\( y \lt \frac{1}{2} \text { and } y \lt -2 \)
\( \therefore \sin x\gt\frac{1}{2} \text { and } \sin x\gt-2 \)
\( \sin x \lt \frac{1}{2} \text { and } \sin x \lt -2\)
\(\therefore x\gt\frac{\pi}{6}\) and \(\sin x\gt-2\)
\(\ldots[\because-1 \leq \sin x \leq 1\), second condition is not possible]
\(\therefore x\gt\frac{\pi}{6}\) and \(\sin x\gt-1 \quad \ldots[\because-1 \leq \sin x \leq 1]\)
\(\therefore x \in\left(\frac{\pi}{6}, \infty\right) \)
Given that \(x^2-x-2 \lt 0 \)
\( \therefore (x-2)(x+1) \lt 0 \)
\( \therefore (x-2) \lt 0 \text { and }(x+1)\gt0 \text { OR } \)
\( (x-2)\gt0 \text { and }(x+1) \lt 0 \)
\( \therefore x \lt 2 \text { and } x\gt-1 \)
\( x\gt2 \text { and } x \lt -1 \)
\( \therefore x \in(-1,2)\)
OR
...(ii) [ \(\because\) second condition is not possible]
\(\therefore\) from (i) and (ii), we get
\(x \in\left(\frac{\pi}{6}, 2\right)\)