Решить:1) cos (6x+1) = -корень из2/22) sinx=2cosx3) ctg (2x- пи/10) = 1/корень из 3

Решите задачу:

$1)\; cos(6x+1)=-\frac{\sqrt2}{2}\\6x+1=\pm arccos(-\frac{\sqrt2}{2})+2\pi n\\6x+1=\pm\frac{\pi}{4}+2\pi n\\6x=\pm\frac{\pi}{4}-1+2\pi n\\x=\pm\frac{\pi}{24}-\frac{1}{6}+\frac{\pi n}{6}, \; n\in Z; \\\\2)\; sinx=2cosx|:cosx, \; cosx \neq 0\\tgx=2\\x=arctg2+\pi n, \; n\in Z;\\\\3)\; ctg(2x-\frac{\pi}{10})=\frac{1}{\sqrt3}\\2x-\frac{\pi}{10}=arcctg\frac{\sqrt3}{3}+\pi n\\2x=\frac{\pi}{3}+\frac{\pi}{10}+\pi n\\x=\frac{1}{2}*(\frac{13\pi}{30}+\pi n)\\x=\frac{13\pi}{60}+\frac{\pi n}{2}, \; n\in Z.$