2sin^8x-2cos^8x=cos^2 2x-cos2x

1. Преобразуем:

2sin^8x - 2cos^8x = cos^2 (2x) - cos2x; 2 (sin^8x - cos^8x) = cos2x (cos2x - 1); 2 (sin^4x + cos^4x) (sin^4x - cos^4x) - cos2x (cos2x - 1) = 0; 2 ((sin^2x + cos^2x) ^2 - 2sin^2xcos^2x) (sin^2x + cos^2x) (sin^2x - cos^2x) + cos2x (1 - cos2x) = 0; - cos2x (2 - sin^2 (2x)) + cos2x (1 - cos2x) = 0; cos2x (1 - cos2x - 2 + sin^2 (2x)) = 0; cos2x (-1 - cos2x + sin^2 (2x)) = 0; cos2x (1 + cos2x - sin^2 (2x)) = 0; cos2x (cos^2 (2x) + cos2x) = 0; cos^2 (2x) (cos2x + 1) = 0.

2. Приравняем множители к нулю:

[cos^2 (2x) = 0;

[cos2x + 1 = 0; [cos2x = 0;

[cos2x = - 1; [2x = π/2 + πk, k ∈ Z;

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

[x = π/2 + πk, k ∈ Z.

Ответ: π/4 + πk/2; π/2 + πk, k ∈ Z.