Question

5cos^2x+7cosx-6=0
Помогите плиз

Answer 1

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