Решите: cos^2x=1/2(1-cosx)

Решите задачу:

$Cos^{2}x=\frac{1-Cosx}{2}\\\\Cos^{2}x-\frac{1-Cosx}{2}=0 \\\\2Cos^{2}x-1+Cosx=0\\\\2Cos^{2}x+Cosx-1=0\\\\Cosx=m,-1 \leq m \leq 1 \\\\2m^{2}+m-1=0\\\\D=1^{2} -4*2*(-1)=1+8=9=3^{2}\\\\m_{1}=\frac{-1+3}{4}=\frac{1}{2} \\\\m_{2} =\frac{-1-3}{4}=-1\\\\Cosx=\frac{1}{2}\\\\x=+-arcCos\frac{1}{2}+2\pi n,nez\\\\x=+-\frac{\pi }{3}+2\pi n,nez\\\\Cosx=-1\\\\x=\pi +2\pi n,nez$