Question

2sin^2-5sinxcosx-cos^2x=-2

Answer 1

$2sin^2x-5sinxcosx-cos^2x=-2 \\ 2sin^2x-5sinxcosx-cos^2x=-2(sin^2x+cos^2x)\\ 2sin^2x-5sinxcosx-cos^2x=-2sin^2x-2cos^2x \\ 4sin^2x-5sinxcosx+cos^2x=0\ \Big| :cos^2x \neq 0 \\ 4tg^2x-5tgx+1=0$
$tg x=1$ или $tgx= \frac{1}{4}$
$x= \frac{ \pi }{4} + \pi k$ или $x= arctg \frac{ 1}{4} + \pi k,\ k \in Z$