1) sin^2x-sinx=0 2) ctg^2x-4ctg x= 0 3) корень с 3 cos2x + sin2x = 0

Решите задачу:

$1) \ sin^2x-sinx=0\\ sinx(sinx-1)=0\\ sinx=0\ \ \ \ \ \ \ \ \ \ sinx=1\\ x=\pi k, \ \ \ \ \ \ \ \ \ \ x=\frac{\pi}{2}+\pi k\ \ \ \ \ \ \ k\ E\ Z \\ x=\frac{\pi}{2}+\frac{\pi k}{2};\ k \ E\ Z.\\ 2)\ ctg^2x-4ctgx=0\\ ctgx(ctgx-4)=0\\ ctgx=0\ \ \ \ \ \ \ \ \ \ \ \ ctgx=4\\ x=\pi k\ \ \ \ \ \ \ \ \ \ \ \ \ \ x=arcctg4+\pi k;\ k\ E\ Z\\ 3)\ \sqrt3cos2x+sin2x=0\\ \frac{\sqrt3}{2}cos2x+\frac{1}{2}sin2x=0\\ cos(\frac{\pi}{6})*cos2x+sin(\frac{\pi}{6})*sin2x=0\\ cos(2x-\frac{\pi}{6})=0\\$
$2x-\frac{\pi}{6}=\frac{\pi}{2}+\pi k;\ k\ E\ Z\\ 2x=\frac{2\pi}{3}+\pi k\\ x=\frac{\pi}{3}+\frac{\pi k}{2}$