Решить уравнения 3cos^3x - sinx - 1 = 0 и 2sin^2 + 3cosx = 0. Tg^2 = 2

Решим уравнения:

1) 3 * cos² x - sin x - 1 = 0;

3 * (1 - sin² x) - sin x - 1 = 0;

Раскроем скобки.

3 - 3 * sin² x - sin x - 1 = 0;

-3 * sin² x - sin x + 2 = 0;

3 * sin² x + sin x - 2 = 0;

D = 1 - 4 * 3 * (-2) = 1 + 12 * 2 = 25 = 5²;

{ sin x = (-1 + 5) / (2 * 6) = 4/12 = 1/3;

sin x = (-1 - 5) / 12 = - 6/12 = - 1/2;

{ sin x = 1/3;

sin x = - 1/2;

{ x = (-1) ^k * arcsin (1/3) + пи * n, n ∈ Z;

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

{ x = (-1) ^k * arcsin (1/3) + пи * n, n ∈ Z;

x = (-1) ^k * 5 пи/6 + пи * n, n ∈ Z.

2) 2 * sin² x + 3 * cos x = 0;

2 * (1 - cos² x) + 3 * cos x = 0;

-2 * cos² x + 3 * cos x + 2 = 0;

2 * cos² x - 3 * cos x - 2 = 0;

D = 9 - 4 * 2 * (-2) = 9 + 16 = 25 = 5²;

{ cos x = (3 + 5) / (2 * 2) = 8/4 = 2 - нет корней;

cos x = (3 - 5) / (2 * 2) = - 2/4 = - 1/2;

cos x = - 1/2;

x = + -2 пи/3 + 2 пи * n, n ∈ Z.

3) tg^2 x = 2;

tg x = + -√2;

{ x = atctg (√2) + пи * n, n ∈ Z;

x = atctg (-√2) + пи * n, n ∈ Z.