11.13. Solve the equation

$$
3-7 \cos ^{2} x \sin x-3 \sin ^{3} x=0
$$
11.13. Answer: $x=\frac{\pi}{2}+2 k \pi$ and $x=(-1)^{k} \frac{\pi}{6}+k \pi$.

Let $t=\sin x$. Considering that $\cos ^{2} x=1-t^{2}$, we obtain the equation $4 t^{3}-7 t+3=0$. This equation has roots $t_{1}=1, t_{2}=1 / 2$ and $t_{3}=-3 / 2$. The last root does not fit.