$2{\sin ^2}x + 5\sin x - 3 = 0$ $\Rightarrow \left( {\sin x + 3} \right)\left( {2\sin x - 1} \right) = 0$ $\sin x = {1 \over 2}$ and $\,\,\sin x \ne - 3$ Given that $x \in \left[ {0,3\pi } \right]$ So possible values of x are $30^\circ$, $150^\circ$, $390^\circ$, $510^\circ$. That means x have 4 values.