Help with a trigonometric equation

2 sin(x)²+3cos(x) = 0

2(1 - cos^2(x)) + 3cos(x) = 0
2cos^2(x) - 3cos(x) - 2 = (2cos(x) + 1)(cos(x) - 2) = 0
cos(x) = -1/2
x = 120 degrees or 240 degrees in [0, 360 degrees)

cos^2(x)+sin^2(x) = 1.. so trade off sin(x)^2 for 1 - cos^2(x)

get 2(1-cos^2(x)) = -3cos(x)

2-2cos^2x = -3cosx
2= -3cosx/-2cos^2x
2=3/2cosx
2*2/3 = 1/cosx
... 1/cosx = secx
therefor
2*2/3 = secx

arcsec(4/3) = x