Cos^2x-5sinx*cosx+2

$cos^2x-5sinxcosx+2(sin^2x+cos^2x)=0 \\ cos^2x+2cos^2x-5sinxcosx+2sin^2x=0 \\ \frac{3cos^2x}{cos^2x}- \frac{5sinxcosx}{cos^2x}+ \frac{2sin^2x}{cos^2x}= \frac{0}{cos^2x} \\ \\ 3-5tgx+2tg^2x =0 \\ 2tg^2x-5tgx+3=0 \\ \\ y=tgx \\ 2y^2-5y+3=0 \\ D=25-24=1 \\ y_{1}= \frac{5-1}{4}=1 \\ \\ y_{2}= \frac{5+1}{4}=1.5$

При у=1
tgx=1
x=π/4 + πk, k∈Z

При у=1,5
tgx=1.5
x=arctg1.5+πk, k∈Z

Ответ: π/4 + πk, k∈Z;
arctg1.5+πk, k∈Z.