Solve for X in this trig question

2 sin^2⁡x-7 sin⁡x+3=0

2sin^2⁡x - 7 sin⁡x + 3 = 0
(2sinx - 1)(sinx - 3) = 0
2sinx - 1 = 0 or sinx - 3 = 0
sinx = 1/2 or sinx = 3
sinx = 3 has no real solutions so discard it, leaving just
sinx = 1/2
x = pi/6 + 2npi or x = 5pi/6 + 2npi for all integers, n

2sin^2x-7sinx+3=0
(2sinx-1)(sinx-3)=0
2sinx-1=0; sinx-3=0
sinx=1/2; sinx=3
π/6, 5π/6, undefined

t = sin(x)

2sin(x)^2 - 7sin(x) + 3 = 0
2t^2 - 7t + 3 = 0
t = (7 +/- sqrt(49 - 4 * 2 * 3)) / (2 * 2)
t = (7 +/- sqrt(49 - 24)) / 4
t = (7 +/- sqrt(25)) / 4
t = (7 +/- 5) / 4
t = 12/4 , 2/4
t = 3 , 1/2
sin(x) = 3 , 1/2

sin(x) can't equal 3

sin(x) = 1/2
x = pi/6 , 5pi/6