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.