2cos^2x+14 cosx=3 sin^2x нужно решить уровнение

[latex]2cos^{2}x+14cosx=3sin^{2}x \\ 2cos^{2}x+14cosx-3sin^{2}x=0 \\ 2cos^{2}x+14cosx-3(1-cos^{2}x)=0\\2cos^{2}x+14cosx-3+3cos^{2}x=0\\5cos^{2}x+14cosx-3=0\\cosx=t\\5t^{2}+14t-3=0\\D=196+60=256\\x_{1}= \frac{-14+16}{10}=0,2 \\ x_{2}= \frac{-14-16}{10}=-3 \\ cosx \neq -3 \\ cosx=0,2\\x_{1}=arccos \frac{1}{5}+2 \pi n\\x_{1}=-arccos \frac{1}{5}+2 \pi n[/latex]