Question

sin2x-9sinxcosx+3cos2x=-1 Прошу помощи.

Answer 1

Sin2x-9sinxcosx+3cos2x=1
2sin(x)cos(x)-9sin(x)cos(x)+3(cos^2(x)-sin^2(x))=cos^2(x)+sin^2(x)
-7sin(x)cos(x)+2cos^2(x)-4sin^2(x)=0
2cos^2(x)-7sin(x)cos(x)-4sin^2(x)=0
-(sin(x)+2cos(x))(4sin(x)-cos(x))=0
Получаем два случая:
sin(x)+2cos(x)=0 => sin(x)/cos(x)=-2 => tg(x)=-2 => x=arctg(-2) +pi*n
и
4sin(x)-cos(x)=0 => sin(x)/cos(x)=1/4 => tg(x)=1/4 => x=arctg(1/4) +pi*n

Answer 2

2sinxcosx-9sincosx+3cos2x+1=0

-7sinxcosx+3cos^2x-3sin^2x+cos^2x+sin^2x=0

4cos^2x-2sin^2x-7sinxcosx=0  :cos^2x

4-2tg^2x-7tgx=0

2s^2+7s-4=0

[-7+-sqrt(49+32)]/4

s1=-4

s2=1/2

x=arctg1/2+Пk

x=arctg(-4)+Пl