Please help me find all the solutions for this equation? cos²x = -2sin x + 2

the answer is π/2 + 2nπ

cos^2(x) = -2sin(x) + 2

use the fact that sin^2(x) + cos^2(x) = 1 ==> cos^2(x) = 1 - sin^2(x)

=> 1 - sin^2(x) = -2sin(x) + 2

rearranging gives

=> sin^2(x) - 2sin(x) + 1 = 0

factorize

=> (sin x - 1)^2 = 0

==>sin x = 1

x = π/2 and (π/2 + 2nπ) where n is any integer

1 - sin²x = -2sin x + 2
sin²x - 2sin x + 1 = (sin x - 1)^2 = 0
sin x = 1
x = π/2 + 2nπ