Решите уравнение 2sin²x - cosx = 1

Решите задачу:

$2Sin^{2}x-Cosx=1\\\\2(1-Cos^{2}x)-Cosx-1=0\\\\2-2Cos^{2}x-Cosx-1=0\\\\2Cos^{2} x+Cosx-1=0\\\\Cosx=m,-1\leq m \leq1\\\\2m^{2}+m-1=0\\\\D=1^{2}-4*2*(-1)=1+8=9=3^{2}\\\\m_{1}=\frac{-1-3}{4}=-1\\\\ m_{2}=\frac{-1+3}{4}=\frac{1}{2} \\\\\left[\begin{array}{ccc}Cosx=-1\\Cosx=\frac{1}{2} \end{array}\right\\\\\left[\begin{array}{ccc}x=\pi+2\pi n,n\in Z \\x=\pm arcCos\frac{1}{2}+2\pi n,n\in Z \end{array}\right$

$\left[\begin{array}{ccc}x=\pi+2\pi n,n\in Z \\x=\pm\frac{\pi }{3}+2\pi n,n\in Z \end{array}\right\\\\Otvet:\boxed{\pi+2\pi n;\pm\frac{\pi }{3}+2\pi n,n\in Z}$