При каких значениях параметра а функция у= 2ax^3 + 9x^2 + 54ax+ 66 убывает на всей числовой прямой

у= 2ax^3 + 9x^2 + 54ax+ 66

найдем производную

  y'=6ax^2+18x+54a

6ax^2+18x+54a< 0

ax^2+3x+9a< 0 

возьмем а=0

3x< 0

x< 0 

d=9-36a=9(1-4a)< =0

1-4a< =0

4a> =1

a> =1/4

0.25 +00 

1cos3x=-2cosx 4cos³x-3cosx+2cosx=0 4cos³x-cosx=0 cosx(4cos²x-1)=0 cosx=0⇒x=π/2+πn,n∈z 4cos²x-1=0 4(1+cos2x)/2=1 1+cos2x=1/2 cos2x=-1/2 2x=+-2π/3+2πk x=+-π/3+πk,k∈z 2 cosx-cos(π/2-11x)=0 -2sin(6x-π/4)sin(-5x+π/4)=0 2sin(6x-π/4)sin(5x-π/4)=0 sin(6x-π/4)=0 6x-π/4=πn 6x=π/4+πn x=π/24+πn/6,n∈z sin(5x-π/4)=0 5x-π/4=πk 5x=π/4+πk x=π/20+πk/5,k∈z 3 (sinx+sin7x)+(cos3x-cos5x)=0 2sin4xcos3x+2sin4xsinx=0 2sin4x*(cos3x+sinx)=0 sin4x=0 4x=πn x=πn/4 cos3x+cos(π/2-x)=0 2cos(x+π/4)cos(2x-π/4)=0 cos(x+π/4)=0 x+π/4=π/2+πk x=π/4+πk,k∈z cos(2x-π/4)=0 2x-π/4=π/2+πm 2x=3π/4+πm x=3π/8+πm/2,m∈z

12(√3/2cosx-1/2sinx)=√2 cos(x+π/6)=√2/2 x+π/6=-π/4+2πn u x+π/6=π/4+2πn x=-5π/12+2πn u x=π/12+2πn,n∈z 2 2(1/2cosx-√3/2sinx)=2cos5x cos(x+π/3)=cos5x 5x=x+π/3+2πn u 5x=-π/3-x+2πn 4x=π/3+2πn u 6x=-π/3+2πn x=π/12+πn/2 u x=-π/18+πn/3,n∈z 3 sin3xcos2x=sin(3x+2x) sin3xcos2x=sin3xcos2x+sin2xcos3x sin2xcos3x=0 sin2x=0⇒2x=πn⇒x=πn/2,n∈z cos3x=0⇒3x=π/2+πn⇒x=π/6+πn/3,n∈z 4 sinxsin7x=sin3xsin5x 1/2[cos(7x-x)-cos(7x+x)]=1/2[cos(5x-3x)-cos(5x+3x)] cos6x-cos8x=cos2x-cos8x cos6x=cos2x 6x=2x+2πn u 6x=-2x+2πn 4x=2πn u 8x=2πn x=πn/2 u x=πn/4-общий