Воспользуемся формулой сложения косинусов:
cosA + cosB = 2cos[(A + B)/2]cos[(A - B)/2]
cosx + cos5x = 0
2cos[(x + 5x)/2]cos[(x - 5x)/2] = 0
2cos3xcos2x = 0
cos3xcos2x = 0
cos3x = 0
3x = π/2 + πn, n ∈ Z
x = π/6 + πn/3, n ∈ Z
2x = π/2 + πk, k ∈ Z
x = π/4 + πk/2, k ∈ Z
Ответ: x = π/4 + πk/2, k ∈ Z; π/6 + πn/3, n ∈ Z.