2cos 7x/2 (cos3x/2+cosx/2)=0 решить,заранее большое !

Решение
2cos 7x/2 (cos3x/2+cosx/2)=0
1) 2cos 7x/2 = 0
cos 7x/2 = 0
7x/2 =  π/2 + πk, k∈Z
7x = π + 2πk, k∈Z
x₁ = π/7 + 2πk/7, k∈Z
2) cos3x/2+cosx/2 = 0
2cos[(3x/2 + x/2) / 2] * cos[(3x/2 - x/2) / 2] = 0
а)  cos[(3x/2 + x/2) / 2] = 0
cosx = 0
x₂ =  π/2 + πn, n∈Z
б) cos[(3x/2 - x/2) / 2] = 0
cosx/2 = 0
x/2 = π/2 + πm, m∈Z
x₃ = π + 2πm, m∈Z