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.
Пошаговое объяснение:
