Помогите решить пожалуйста. 1) 2sin(2-п/6)=1  2) cosx (cos2x-1)=0

[latex]2sin(2x-\frac{\pi}{6})=1\\sin(2x-\frac{\pi}{6})=\frac{1}{2}\\2x-\frac{\pi}{6}=(-1)^narcsin\frac{1}{2}+\pi n\\ 2x-\frac{\pi}{6}=(-1)^2\frac{\pi}{6}+\pi n\\2x=(-1)^n\frac{\pi}{3}+\pi n\\x=(-1)^n\frac{\pi}{6}+ \frac{\pi n}{2}, \; n\in Z; \\\\cosx(cos2x-1)=0\\1)cosx=0\\x=\frac{\pi}{2}+\pi n, \; n\in Z; \\\\2)cos2x-1=0\\cos2x=1\\2x=2\pi n\\x=\pi n, \; n\in Z[/latex]