5cos^2x+7cosx-6=0 Помогите плиз
5 cos^2 x + 7 cos x - 6 = 0; cos x = t; -1 ≤ t ≤ 1; 5 t^2 + 7t - 6 = 0; D = 7^2 - 4*5*(-6) = 49 + 120 = 169 = 13^2; t1 = (-7-13)/10 = - 2 < - 1; нет решений или t2 = (-7+13) / 10 = 3/5; cos x = 3/5; x = +- arccos(3/5) + 2pi*k; k-Z