Solve 2cosx-sin^x +2 =0 for all real values of x

Answer 1

You must have a typo.
I will assume you meant ...
2cosx-sin^2 x +2 =0

2cosx - (1 - cos^2x) + 2 = 0
cos^2x + 2cosx + 1 = 0
(cosx + 1)^2 = 0
cosx + 1 = 0
cosx = -1
x = 180º or x = pi radians