Question

cos6x-cos3x=0 на отрезке[0.п]

с разъяснением

Answer 1

cos^2(3x)-(1-cos^2(3x))-cos3x=0

2*cos^2(3x)-cos(3x) -1=0

cos(3x)=t

2t^2-t-1=0

t=1

t=-0.5

1)cos3x=1

3x=2πn

x=(2πn)/3 (n∈Ζ)

2) cos3x=-0.5

3x=2π/3+2πk ==> x=2π/9+2πk/3 (k∈Ζ).

3x=4π/3+2πl ==>  x=4π/3+2πl/3 (l∈Ζ)