Cos(6x) - cos(3x) = -2*sin(9x/2)*sin(3x/2) = 0
sin(9x/2) = 0, 9x/2=πk, x=2πk/9
sin(3x/2) = 0, 3x/2=πk, x=2πk/3
Найдем, при каких к корни будут принадлежать указанному промежутку:
0≤2πk/9≤π, 0≤k≤4.5 - т.е. k=0, 1, 2, 3, 4
0≤2πk/3≤π, 0≤k≤1.5 - т.е. k=0, 1
x∈[0;π]
k=0, x=0
k=1, x=2π/9, x=2π/3
k=2, x=4π/9, x=4π/3
k=3, x=6π/9 = 2π/3, x=6π/3 = 2π
k=4, x=8π/9
Ответ: 0, 4π/9, 2π/3, 8π/9