алгебра - Решите тригонометрическое уравнение

$$\sin \left ( \frac {4x}{3}+\frac {\pi}{6} \right )=- \frac {1}{2}$$ $$\frac {4x}{3}+\frac {\pi}{6}=\left ( -1 \right )^k\arcsin \left ( - \frac 12 \right )+\pi k$$ $$\frac {4x}{3}+\frac {\pi}{6}=\left ( -1 \right )^{k+1}\arcsin \left ( \frac 12 \right )+\pi k$$ $$\frac {4x}{3}+\frac {\pi}{6}=\left ( -1 \right )^{k+1} \frac {\pi}{6} +\pi k$$ $$\frac {4x}{3}=-\frac {\pi}{6}+\left ( -1 \right )^{k+1} \frac {\pi}{6} +\pi k$$ $$x=-\frac {\pi}{8}+\left ( -1 \right )^{k+1} \frac {\pi}{8} +\frac {3\pi k}{4}$$