Find all solutions of the equation in the interval 0 2pi 2

Find all solutions of the equation in the interval [ 0, 2pi )

2sin²x = 2 - cosx

Write your answer in radians in terms of pi. If there is more than one solution, separate them with commas.

2sin^2x = 2 - cosx

as we know sin^2x +cos^2x =1

So, 2( 1-cos^2x) = 2 -cosx

2 -2cos^2x = 2 -cosx

-2cos^2x +cosx =0

cosx( 1 -2cosx) =0

cosx =0

x= pi/2, 3pi/2

1-2cosx =0 ; cosx =1/2

x= cos^-1(1/2) = pi/3, 2pi-pi/3

= pi/3 , 5pi/3

Solution in the interval [ 0, 2pi ) : x= pi/2 , pi/3 , 3pi/2 , 5pi/3