Решите задание 3 пожалуйста

Cosx≠1/2, cosx≠0
$\frac{4cosx-5}{2cosx-1} + \frac{1}{2cos^2x-cosx} =2 \\ \frac{4cosx-5}{2cosx-1} + \frac{1}{cosx (2cosx-1)} =2 \\ \frac{cosx(4cosx-5)}{cosx (2cosx-1)} + \frac{1}{cosx (2cosx-1)} =2 \\ \frac{cosx(4cosx-5)+1}{cosx (2cosx-1)} =2$
cosx(4cosx-5)+1 =2cosx (2cosx-1)
4cos²x-5cosx+1=4cos²x-2cosx
5cosx-2cosx=1
3cosx=1
cosx=1/3
x=+-arccos (1/3)+2πn
б)
х=+arccos (1/3)-4π