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Kubns
@Kubns
July 2022
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6sin^2(x)-sin(x)-1≤0
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zarembo73
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;
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Answers & Comments
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;