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cosx+cos2x=0

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