Решите 2sin^2x+5sinx cosx+cos^2x-4=0 )

2sin^2x+5sinxcosx+cos^2x-4(sin^2x+cos^2x)=0
2sin^2x+5sinxcosx+cos^2x-4sin^2x-4cos^2x=0
-2sin^2x+5sinxcosx-3cos^2x=0
2sin^2x-5sinxcosx+3cos^2x=0
2tg^2x-5tgx+3=0
[ tgx=1
tgx=2/3
x=pi/4+ pi*n, nєZ
x=arctg 2/3 +pi*k, kє Z