Solve for $x$ in $4 \sin ^{2} x+3 \sin x+2=-3 \sin x$ $x=0$ $x=-\frac{\pi}{3}$ $x=\frac{5 \pi}{6}$ $x=\frac{2 \pi}{3}, x=\pi$ $x=-\frac{\pi}{6}, x=-\frac{\pi}{2}$

Step 1 ：The given equation is a quadratic equation in terms of \(\sin x\). We can solve it by setting it equal to zero and then using the quadratic formula. The solutions will be the values of \(\sin x\) that satisfy the equation. We can then find the corresponding values of \(x\) by using the inverse sine function.

Step 2 ：The solutions provided are \(x = -\frac{\pi}{2}\), \(x = -\frac{\pi}{6}\), \(x = \frac{7\pi}{6}\), and \(x = \frac{3\pi}{2}\). However, these solutions need to be checked against the original equation to ensure they are valid.

Step 3 ：The solutions have been checked and found that all of them are valid. Therefore, the solutions to the equation are \(x = -\frac{\pi}{2}\), \(x = -\frac{\pi}{6}\), \(x = \frac{7\pi}{6}\), and \(x = \frac{3\pi}{2}\).

Step 4 ：\(\boxed{\text{Final Answer: The solutions to the equation are } x = -\frac{\pi}{2}, x = -\frac{\pi}{6}, x = \frac{7\pi}{6}, \text{ and } x = \frac{3\pi}{2}}\)