Question

Помогите
8sin2x+cosx+1=0

Answer 1

Итак, 
8*(1-cos^2x)+cosx+1=0

8-8cos^2x+cosx+1=0

-8cos^2x+cosx+9=0

8cos^2x-cosx-9=0

cosx=y

8y^2-y-9=0

Answer 2

$8sin^2x+cos x+1=0\8(1-cos^2x)+cos x+1=0\8cos^2x+cos x+1=0\cos^2x-cos x-9=0\cos x=t,;cos^2x=t^2,;tin[-1;;1]\8t^2-t-9=0\D=1+4cdot8cdot9=1+288=289=(17)^2\t_1=frac{1-17}{16}=-1\t_2=frac{1+17}{16}=frac{18}{16}>1;-;HE;nogx.\cos x=-1\x=pi+2pi n,;ninmathbb{Z}$