Can anyone help plz? Find all solutions of the equation in the interval (0,2π)2sin^2x-3cosx=3 Write your answer in radians in terms of π?

2 [ 1 - cos²x ] - 3 cos x = 3
2 cos²x + 3 cos x + 1 = 0
[ 2 cos x + 1 ] [ cos x + 1 ] = 0
cos x = - 1/2 , cos x = - 1
x = 2π/3 , 4π/3 , π

Hint: sin² x = 1 - cos² x

use that substitution to get a quadratic equation...

2 - 2 cos² x - 3 cos x = 3

solve: 2 cos² x + 3 cos x + 1 = 0

(2 cos x + 1)(cos x + 1) = 0

cos x = - 1/2 or cos x = - 1

x = 2π/3, π, 4π/3

check...do something !!