Решите уравнение: 6sin^2 х-cos x-5=0

6(1 - cos²x) - cosx - 5 =0;  6cos²x + cosx - 1 =0;  cosx = a;  6a² +a - 1 = 0; a = - 1/2; a = 1/3; cosx = +-2π/3 +2πn,n∈Z;  cosx = 1/3; x = +- arccos1/3 +2πn,n∈Z;