решить уравнение: cos^2 x + cos^2 2x + cos^2 3x = 3/2

1. Умножим обе части на 2:

cos^2 (x) + cos^2 (2x) + cos^2 (3x) = 3/2; 2cos^2 (x) + 2cos^2 (2x) + 2cos^2 (3x) = 3; 2cos^2 (x) - 1 + 2cos^2 (2x) - 1 + 2cos^2 (3x) - 1 = 0.

2. Косинус двойного угла и сумма косинусов:

cos2x + cos4x + cos6x = 0; 2cos ((6x + 2x) / 2) * cos ((6x - 2x) / 2) + cos4x = 0; 2cos4x * cos2x + cos4x = 0; cos4x (2cos2x + 1) = 0; [cos4x = 0;

[2cos2x + 1 = 0; [cos4x = 0;

[2cos2x = - 1; [cos4x = 0;

[cos2x = - 1/2; [4x = π/2 + πk, k ∈ Z;

[2x = ±2π/3 + 2πk, k ∈ Z; [x = π/8 + πk/4, k ∈ Z;

[x = ±π/3 + πk, k ∈ Z.

Ответ: π/8 + πk/4; ±π/3 + πk, k ∈ Z.