Sin^2x+cos^2 2x+sin^2 3x=3/2

1. Косинус двойного угла:

sin^2x + cos^2 (2x) + sin^2 (3x) = 3/2; 2sin^2x + 2cos^2 (2x) + 2sin^2 (3x) = 3; 2sin^2x - 1 + 2cos^2 (2x) - 1 + 2sin^2 (3x) - 1 = 0; - (1 - 2sin^2x) + (2cos^2 (2x) - 1) - (1 - 2sin^2 (3x)) = 0; - cos2x + cos4x - cos6x = 0.

2. Сумма косинусов:

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

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

[cos2x = 1/2; [4x = π/2 + πk, k ∈ Z;

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

[x = ±π/6 + πk, k ∈ Z.

Ответ: π/8 + πk/4, k ∈ Z; ±π/6 + πk, k ∈ Z.