6sin^2 (x) - sin (x) - 1≤0

6sin² x-sinx-1≤0;

sinx=t, - 1 ≤t≤1;

6t²-t-1≤0;

D=1+24=25;

t1 = (1-5) / 12=-4/12=-1/3∈[-1; 1];

t2 = (1+5) / 12=6/12=1/2∈[-1; 1];

6 (t+1/3) (t-1/2) ≤0;

(t+1/3) (t-1/2) ≤0;

-1/3≤t≤1/2;

-1/3≤sinx≤1/2;

sinx≥-1/3;

arcsin (-1/3) + 2πn≤x≤π-arcsin (-1/3) + 2πn, n∈Z;

-arcsin1/3+2πn≤x≤π+arcsin1/3+2πn, n∈Z;

sinx≤1/2;

-π-arcsin1/2+2πn≤x≤arcsin1/2+2πn, n∈Z;

-π-π/6+2πn≤x≤π/6+2πn, n∈Z;

-7π/6+2πn≤x≤π/6+2πn, n∈Z.

Ответ: [-arcsin1/3+2πn; π+arcsin1/3+2πn]∪[-7π/6+2πn; π/6+2πn], n∈Z;