Решите уравнение 2sin^2x-cos2x=1

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

Упростим уравнение.

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

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

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

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

3 * sin^2 x - (1 - sin^2 x) = 1;

3 * sin^2 x - 1 + sin^2 x = 1;

4 * sin^2 x - 1 = 1;

4 * sin^2 x = 1 + 1;

4 * sin^2 x = 2;

sin^2 x = 2/4;

sin^2 x = 1/2;

Получим корни.

1) sin x = 1/2;

x = (-1) ^n * arcsin (1/2) + пи * n, n ∈ Z;

x = (-1) ^n * пи/6 + пи * n, n ∈ Z;

2) sin x = - 1/2;

x = (-1) ^n * arcsin (-1/2) + пи * n, n ∈ Z;

x = (-1) ^n * 5 * пи/6 + пи * n, n ∈ Z.