Решите уравнение6sin²x + 5cosx - 7 = 0

6 * sin² x + 5 * cos x - 7 = 0;

Вычислим корни.

6 * (1 - cos² x) + 5 * cos x - 7 = 0;

6 * 1 - 6 * cos² x + 5 * cos x - 7 = 0;

-6 * cos² x + 5 * cos x - 7 + 6 = 0;

-6 * cos² x + 5 * cos x - 1 = 0;

6 * cos² x - 5 * cos x + 1 = 0;

Пусть cos x = a, a ∈ [-1; 1].

6 * a² - 5 * a + 1 = 0;

D = (-5) ² - 4 * 6 * 1 = 25 - 24 = 1;

a1 = (5 + 1) / (2 * 6) = 6/12 = 1/2;

a2 = (5 - 1) / 12 = 4/12 = 1/3;

1) cos x = 1/2;

x = + -arccos (1/2) + п * n, n ∈ Z;

x = + -п/3 + п * n, n ∈ Z;

2) cos x = 1/2;

x = + -arccos (1/3) + п * n, n ∈ Z.