Решить уравнение 2cos^2x+2cosx=3sin^2x

Решите задачу:

$2cos^{2} x+2cosx=3sin^{2} x \\ 2cos^{2} x+2cosx-3sin^{2} x =0 \\ 2cos^{2} x+2cosx-3(1-cos^{2} x) =0 \\ 2cos^{2} x+2cosx-3+3cos^{2} x =0 \\ 5cos^{2} x+2cosx-3=0 \\ cosx=t \\ 5t^{2} +2t-3=0 \\ D=4+60=64 \\ \sqrt{D} =8 \\ t _{1} = \frac{-2+8}{10} =0.6 \\ t _{2} = \frac{-2-8}{10} =-1 \\ cosx=0.6 \\ x=+-arccos0.6+2 \pi n \\$ $cosx=-1 \\ x= \pi +2 \pi m$