Solve 2sin²(x) = 1 - sin (x)

find all the values of X in degrees ( 0 < X < 360)
show work please

2sin²(x) = 1 - sin(x)

2sin²(x) + sin(x) - 1 = 0

Let y = sin(x)

Use:
2y² + y - 1 = 0

Factor:
(2y - 1)(y + 1) = 0

y = 1/2
or y = -1

sine values are positive in Quadrant 1 and 2, and negative in Quadrants 3 and 4.

sin(x) = 1/2
or sin(x) = -1

sin(x) = 1/2
π/6 which is 30° and 5π/6 which is 150°

sin(x) = -1
3π/2 or 270°