Решите уравнение: (cos^2)(2x)+(cos^2)(4x)=1+cos(8x)

  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: