Решите уравнение: cosx + cos2x + cos3x = 0 на отрезке .(o; 2пи)

1. cosx + cos2x + cos3x = 0

cos2x + (cosx + cos3x) = 0

cos2x + 2 cos (3x + x / 2) cos (3x - x / 2) = 0

cos2x + 2 cos2xcosx = 0

cos2x (1 + 2cosx) = 0

cos2x = 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 = 0

cos x = -1/2

x1 = 2π/3 + 2πn;

x2 = -2π/3 + 2πn;

n принадлежит Z.

2. на отр-ке (0;2π): 

1) π/4, 3π/4, 5π/4, 7π/4

2) 2π/3, 4π/3