The number of values of x in the interval [0,3 π] satisfying the...

Solving $2\sin^2x + 5\sin x - 3 = 0$

The equation is quadratic in $\sin x$. Substitute $y = \sin x$, yielding $2y^2 + 5y - 3 = 0$.

Factoring gives $(2y - 1)(y + 3) = 0$. Solutions are $y = \frac{1}{2}$ and $y = -3$.

Since the range of sine is $[-1, 1]$, $\sin x = -3$ has no solution.

For $\sin x = \frac{1}{2}$, the general solution is $x = n\pi + (-1)^n\frac{\pi}{6}$, where $n \in \mathbb{Z}$.

Within $[0, 3\pi]$, valid values occur for $n = 0, 1, 2, 3$, giving $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$.

Thus, there are exactly four solutions.

$\boxed{4}$