Given equation is

\( \Large 2\sin^{2}x+3 \sin x - 2 > 0 \)

\( \Large \left(2\sin x -1\right) \left(\sin x + 2\right) > 0 \)

=> \( \Large 2 \sin x -1 > 0 \)

=> \( \Large \sin x > \frac{1}{2} => x \leftarrow \left(\frac{ \pi }{6}, \frac{5 \pi }{6}\right) \)..(i)

Also, \( \Large x^{2}-x-2<0 \)

=> \( \Large \left(x-2\right) \left(x+1\right)<0 => -1 < x < 2 \) .. (ii)
from equation (i) and (ii). we get that x must lies in \( \Large \left(\frac{ \pi }{6}, 2\right) \)