Решите уравнения cos2x=cosx 2sin^2x = cosx-1

Найдем корни.

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

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

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

-2 * sin (x/2) * sin (3 * x/2) = 0;

{ sin (x/2) = 0;

sin (3 * x/2) = 0;

{ x/2 = пи * n, n ∈ Z;

3 * x/2 = пи * n, n ∈ Z;

{ x = 2 * пи * n, n ∈ Z;

3 * x = 2 * пи * n, n ∈ Z;

{ x = 2 * пи * n, n ∈ Z;

x = 2 * пи/3 * n, n ∈ Z.

2) 2 * sin^2 x = cos x - 1;

2 * sin^2 x - cos x + 1 = 0;

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

-2 * cos^2 x - cos x + 3 = 0;

2 * cos^2 x + cos x - 3 = 0;

D = 1 - 4 * 2 * (-3) = 25 = 5;

cos x = (-1 + 5) / (2 * 2) = 4/4 = 1;

cos x = (-1 - 5) / 4 = - 6/4;

1. cos x = - 6/4;

Нет корней.

2. cos x = 1;

x = 2 * пи * n, n ∈ Z.