4. Denoting $\operatorname{Sin} x=a, \operatorname{Sin} 2 x=b, \operatorname{Sin} 3 x=c$, we get $a^{3}+b^{3}+c^{3}=(a+b+c)^{3}$. This equality can easily be reduced to the form: $(a+b)(a+c)(b+c)=0$. Then, using the formula $\operatorname{Sin} \lambda+\operatorname{Sin} \beta=2 \operatorname{Sin} \frac{\lambda+\beta}{2} \operatorname{Cos} \frac{\lambda-\beta}{2}$, we obtain three series of solutions:
Answer: $\left\{\frac{2 k \pi}{3}, \frac{2 m \pi}{5}, \frac{\pi n}{2}\right\} k, m, n \in Z$.