Solve for x in the interval [0,2pi) sin^2x+2cosx=2

Answer 1

sin^2x+2cosx=2
1 - cos^2x + 2cosx - 2 = 0
cos^2x - 2cosx + 1 = 0
(cosx - 1)^2 = 0
.
.
.
cosx = 1

Can you take it from there?