Решите уравнение cosx+cos5x+2sin^2 = 1

cos x + cos (5 * x) + 2 * sin^2 х = 1;

Упростим уравнение.

cos x + cos (5 * x) + 2 * sin^2 х - 1 = 0;

cos x + cos (5 * x) + 2 * sin^2 х - (sin^2 x + cos^2 x) = 0;

2 * cos ((x + 5 * x) / 2) * cos ((x - 5 * x) / 2) + 2 * sin^2 x - sin^2 x - cos^2 x = 0;

2 * cos (6 * x/2) * cos (-4 * x/2) + sin^2 x - cos^2 x = 0;

2 * cos (3 * x) * cos (-2 * x) - (cos^2 x - sin^2 x) = 0;

2 * cos (3 * x) * cos (2 * x) - cos (2 * x) = 0;

cos (2 * x) * (2 * cos (3 * x) - 1) = 0;

1) cos (2 * x) = 0;

2 * x = 2 * пи * n, n ∈ Z;

x = пи * n, n ∈ Z;

2) 2 * cos (3 * x) = 1;

cos (3 * x) = 1/2;

3 * x = + -пи/3 + 2 * пи * n, n ∈ Z;

x = + -пи/9 + 2 * пи/3 * n, n ∈ Z.