Step 1: Treat it as a quadratic.
The equation $\sin^2 x - \sin x - 2 = 0$ is a quadratic if we let $t = \sin x$. So it becomes $t^2 - t - 2 = 0$, which we can factorise.
Step 2: Factorise.
Find two numbers multiplying to $-2$ and adding to $-1$. They are $-2$ and $+1$.
\[ (t - 2)(t + 1) = 0 \] So $t = 2$ or $t = -1$.
Step 3: Check the first root.
Recall $\sin x$ can only lie between $-1$ and $1$. The value $t = 2$ means $\sin x = 2$, which is impossible. So this gives no solution.
Step 4: Check the second root.
The value $t = -1$ means $\sin x = -1$, which is allowed.
Step 5: Count solutions in the interval.
In the range $[0, 2\pi]$, the sine equals $-1$ only at $x = \frac{3\pi}{2}$. That is just one point.
Step 6: State the count.
So there is exactly $1$ solution.
\[ \boxed{1} \]