Геометрия: 2cos^{2}(x+270)=3sin(x+270)

Решите задачу:

$2cos^2(x+270)=3sin(x+270) \\\ 2cos^2(x+ \frac{3\pi}{2})=3sin(x+\frac{3\pi}{2} ) \\\ 2cos^2y=3siny \\\ 2-2sin^2y-3siny=0 \\\ 2sin^2y+3siny-2=0 \\\ D=9+16=25 \\\ siny_1 \neq -2<-1 \\\ siny_2=0.5 \\\ y= (-1)^k \frac{\pi}{6} +\pi k \\\ x+ \frac{3\pi}{2}= (-1)^k \frac{\pi}{6} +\pi k \\\ x= (-1)^k \frac{\pi}{6} +\pi k- \frac{3\pi}{2}, k\in Z$