Solve for 0 <= 0 <= 2pi : cos^2x = 1 + sinx?

Question

My answer key says the answers are:

5pi/6, pi, 7pi/6, 2pi

how do I get those answers?

Answer 1

1 - 2 sin ² x = 1 + sin x
2 sin ² x + sin x = 0
sin x (sin x + 1) = 0
sin x = 0 , sin x = - 1
x = 0 , π , 2π , 3π/2

Answer 2

cos^2x = 1-sin^2x = (1+sinx)(1-sinx)

For sinx <> -1, cos^2x = 1+sinx means 1 - sinx = 1 or sinx = 0, which means x = 0 or pi

If sinx = -1 then x = 3pi/2 and cos^2x = cos^2(3pi/2) = 0 (which is also a solution of the equation).

Therefore the three solutions are x = 0, pi, 3pi/2

Answer 3

1- sin^2x = 1+ sinx
sin^2x + sinx = 0
sinx(sinx + 1) = 0
sinx = 0 x = 0 or pi
sinx = -1 x = 3pi/2

Answer 4

1 - sin^2 x = 1 + sinx
So sin^2 x = -sinx

This can only happen if sin(x) = 0 or -1
or x = 0, π, 3π/2, 2π

*EDIT*
The answers you provided do not match in the original equation. Check it yourself.