Решить уравнение нужно. 5tgx-8ctg+6=0 sin2x+1=4cos^x 14cos^x+3=3cos^x-10sin^x )

5tgx-8ctgx+6=0   |*tgx

5tg^2x+6tgx-8=0

tgx=t

5t^2+6t-8=0

d=36-4*5*(-8)=196

t=-2

t=4/5

1)tgx=-2

x=arctg(-2)+pik . k=z

2)tgx=4/5

x=arctg(4/5)+pik . k=z

sin2x+1=4cos^2x

2sinxcosx=4cos^2x-1   | /cos^2x

2tgx=4-1/cos^2x

-2tgx=(1/cos^2x   -1) -3

-2tgx=tg^2x-3

tg^2x+2tgx-3=0

tgx=t

t^2+2t-3=0

t=1

t=-3

1)tgx=1

x=pi/4+pik . k=z

2)tgx=-3

x=arctg(-3)+pik . k=z

14cos^2x+3=3cos^2x-10sin^2x  

11cos^2x+10sin^2x+3=0

10+cos^2x+3=0

cosx^2=-13  -нет корней

(m+3)^3-8 = (m+3)^3 - 2^3 = (m+3-+3)²+2(m+3)+4) =

= (m+1)(m²+6m+9+2m+6+4)=(m+1)(m²+8m+19);

(x^2+1)^2-4x^2 = (х²+1-2х)(х²+1+2х)=(х-1)²(х+1)²

b1 = b2/q = -6

ответ: -6