Сosx + cos2x + cos3x = 0

1. cosx + cos2x + cos3x = 0cos2x + (cosx + cos3x) = 0cos2x + 2 cos (3x + x / 2) cos (3x - x / 2) = 0cos2x + 2 cos2xcosx = 0cos2x (1 + 2cosx) = 0cos2x = 0 или 1 + 2cosx = 0 1) cos2x = 0 cos2x = cosπ/2 + πn 2x = π/2 + πn x = π/4 + πn/2; n принадлежит Z. 2) 1 + 2 cosx = 0cos x = - 1/2 x1 = 2π/3 + 2πn; x2 = - 2π/3 + 2πn; n принадлежит Z.

(cosx+cos3x) + cos2x=0

2cos2x*cosx+cos2x=0

cos2x (2cosx+1) = 0

cos2x=0

2x=±arccos 0 + 2πn, n ∈ Z

2x=±π/2+πn, n ∈ Z

x=π/4+πn/2, n ∈ Z

2+cosx = 0

2cosx+1=0

cosx=-1/2

x=±2π/3 + 2πn, n ∈ Z

Ответ: π/4+πn/2, ±2π/3 + 2πn