3cos^2x+5sinx cosx=-1

Решите задачу:

$3cos^2x+5sinxcosx=-1\\3cos^2x+5sinxcosx+1=0\\3cos^2x+5sinxcosx+sin^2x+cos^2x=0\\sin^2x+5sinxcosx+4cos^2x=0|:cos^2x\\tg^2x+5tgx+4=0\\tgx=t,t\in R\\t^2+5t+4=0\\D=25-4*4=25-16=9\\\\t_1=\frac{-5+3}2=-1\\\\t_2=\frac{-5-3}2=-4\\\\ \left[\begin{array}{ccc}tgx=-1\\tgx=-4\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}x=-\frac{\pi}4+\pi n;n\in Z\\x=-arctg(4)+\pi n;n\in Z\end{array}\right$