Решить уровнение 2cos(3x-pi/6)=-1

[latex]2cos(3x-\frac{\pi}{6})=-1\\\\cos(3x-\frac{\pi}{6})=-\frac{1}{2}\\\\3x-\frac{\pi}{6}=\pm (\pi -\frac{\pi}{3})+2\pi n=\pm \frac{2\pi}{3}+2\pi n,\; n\in Z\\\\3x=\pm \frac{2\pi}{3}+\frac{\pi}{6}+2\pi n,\; n\in Z\\\\x=\pm \frac{2\pi}{9}+\frac{\pi}{18}+\frac{2\pi n}{3}= \left [ {{\frac{5\pi }{18}+\frac{2\pi n}{3},\; n\in Z} \atop {-\frac{\pi}{6}+\frac{2\pi k}{3},\; k\in Z}} \right. [/latex]