The number of values of $x$ in the interval $[0,3\pi ]$ satisfying the equation $2{\sin }^{2}x+5\sin x-3=0$ is:

Given equation: $2{\sin }^{2}x+5\sin x-3=0$ Let $\sin x=t$. Then $2t^2+5t-3=0$.Factorizing:
$$(2t-1)(t+3)=0$$
So, $t=1/2$ or $t=-3$.Since $-1\le \sin x\le 1$, $t=-3$ is impossible.Therefore, $\sin x=1/2$.In the interval $[0,3\pi ]$:In $[0,\pi ]$, $\sin x=1/2$ at $x=\pi /6$ and $x=5\pi /6$.In $[\pi ,2\pi ]$, $\sin x$ is negative, so no values.In $[2\pi ,3\pi ]$, $\sin x=1/2$ at $x=2\pi +\pi /6$ and $x=2\pi +5\pi /6$.Total values = $4$.