Can't figure it out!! 2sin^2x+3cosx-3=0 ?

Question

I've been on this problem all summer and have not yet figured it out

solve for x and all solutions have to be between 0 and 2 pi

Hint: sin^2x=1-cos^2x

Thanks

Answer 1

2sin² x + 3cosx - 3 = 0

=> 2(1 - cos² x) + 3cosx - 3 = 0

=> 2cos² x -3cosx + 1 = 0

=> (2cosx - 1)(cosx - 1) = 0

=> 2cosx - 1 = 0 or cosx - 1 =0

=> cosx = 1/2 or cosx = 1

=> x = π/3 or 5π/3 or x = 0 or 2π

=> x = { 0, π/3, 5π/3, 2π }.