Решить уравнение: cosx+cos2x+cos6x+cos7x=0

1. Воспользуемся формулой:

cosx + cosy = 2cos ((x + y) / 2) * cos ((x - y) / 2); cosx + cos2x + cos6x + cos7x = 0; 2cos ((7x + x) / 2) cos ((7x - x) / 2) + 2cos ((6x + 2x) / 2) cos ((6x - 2x) / 2) = 0; 2cos4x * cos3x + 2cos4x * cos2x = 0; 2cos4x (cos3x + cos2x) = 0; 2cos4x * 2cos ((3x + 2x) / 2) cos ((3x - 2x) / 2) = 0; 2cos4x * 2cos (5x/2) * cos (x/2) = 0; cos4x * cos (5x/2) * cos (x/2) = 0.

2. Приравняем к нулю множители:

[cos4x = 0;

[cos (5x/2) = 0;

[cos (x/2) = 0; [4x = π/2 + πk, k ∈ Z;

[5x/2 = π/2 + πk, k ∈ Z;

[x/2 = π/2 + πk, k ∈ Z; [x = π/8 + πk/4, k ∈ Z;

[x = π/5 + 2πk/5, k ∈ Z;

[x = π + 2πk, k ∈ Z.

Ответ: π/8 + πk/4; π/5 + 2πk/5; π + 2πk, k ∈ Z.