Решение
cosx+cos2x=0
cosx + 2cos^2x - 1 = 0
2cos^2x + cosx - 1 = 0
D = 1 + 4*2*1 = 9
1) cosx = (-1 -3)/4
cosx = - 1
x1 = π + 2πn, n∈Z
2) cosx = (-1+3)/4
cosx = 1/2
x = (+ -)arccos(1/2) + 2πk, k∈Z
x2 = (+ -)π/3 + 2πk, k∈Z
Ответ: x1 = π + 2πn, n∈Z; x2 = (+ -)π/3 + 2πk, k∈Z