Solve on the interval [0,2pi) 2sin^2x-3sinx+1=0

2sin² (x) + 3sin (x) + 1 = 0

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

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

sin (x) = - 1

x = {3π/2}

2sin (x) + 1 = 0

sin (x) = - 1/2

x = {7π/6, 11π/6}

x = {7π/6, 3π/2, 11π/6}