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

The equation $2\sin^2 x + 5\sin x - 3 = 0$ can be treated as a quadratic in $\sin x$. Substituting $y = \sin x$ yields $2y^2 + 5y - 3 = 0$, which factors as $(2y - 1)(y + 3) = 0$.

This gives $y = \frac{1}{2}$ or $y = -3$. Since $\sin x$ must lie in $[-1, 1]$, only $\sin x = \frac{1}{2}$ is valid.

The general solution is $x = n\pi + (-1)^n \frac{\pi}{6}$ for integer $n$. In $[0, 3\pi]$, the valid values occur for $n = 0, 1, 2, 3$, giving $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$.

The total count of solutions is $\boxed{4}$.