Solution: Let $ y = \sin(2x) $. The equation becomes $ y^2 + 3y + 2 = 0 $, which factors as $ (y + 1)(y + 2) = 0 $. Thus, $ y = -1 $ or $ y = -2 $. Since $ \sin(2x) $ ranges between $-1$ and $1$, $ y = -2 $ is invalid. For $ y = -1 $, $ \sin(2x) = -1 $ has infinitely many solutions (e.g., $ 2x = rac3\pi2 + 2\pi k $, $ k \in \mathbbZ $). However, if restricted to a specific interval (not stated here), the count would depend on the domain. Assuming $ x \in \mathbbR $, there are infinitely many solutions. But if the question implies a general count, the answer is oxed extinfinitely many. - Crosslake