find all solutions of equation from (0, 2pi) work and steps please?

Question

1. 2sin^2x = 2 + cos x

work and steps please

Answer 1

Since sin^2 x + cos^2 x = 1, sin^2 x = 1 - cos^2 x.
Replace the sin^2x with 1 - cos^2 like this:

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

That's 2 - 2 cos^2 x = 2 + cosx

Move every thing now on the left to the right by adding their opposite: the 2 will go away leaving

0 = 2 cos^2 x + cos x

Factor out the cos x: 0 = cos x(2 cos x + 1)

So either cos x = 0 meaning x = pi/2 or 3 pi/2

Or, 2 cosx + 1 = 0 meaning cos x = -1/2 which in turn means x is in quadrant II or III and equivalent to pi/3 (since cos pi/3 = +1/2)

So 2 pi / 3 or 4 pi / 3

Those 4 places are the solution set.

Answer 2

2(1 - cos² x) = 2 + cos x
2 - 2.cos² x = 2 + cos x
2.cos² x + cos x = 0
cos x.(2 cos x + 1) = 0
cos x = 0 , cos x = - 1/2
x = π/2 , 3π/2 , 2π/3 , 4π/3