3 sin t >= корень из 2 на 2...

August 2022 1 18 Report

cos t>=1/3
cos t <= -1/2
sin t<=-1/2
sin t> корень из 3 на 2
sin t <=2/3
sin t >= корень из 2 на 2

Answers & Comments

zarembo73 1) cos t≥1/3;
-arccos 1/3+2πn≤t≤arccos 1/3+2πn, n∈Z.

2) cos t≤-1/2 ;
arccos (-1/2)+2πn≤t≤2π-arccos (-1/2)+2πn, n∈Z;
π-arccos 1/2+2πn≤t≤2π-(π-arccos 1/2)+2πn, n∈Z;
π-π/3+2πn≤t≤π+π/3+2πn, n∈Z;
2π/3+2πn≤t≤4π/3+2πn, n∈Z.

3) sin t≤-1/2;
-π-arcsin(-1/2)+2πn≤t≤arcsin(-1/2)+2πn, n∈Z;
-π+arcsin 1/2+2πn≤t≤-arcsin 1/2+2πn, n∈Z;
-π+π/6+2πn≤t≤-π/6+2πn, n∈Z;
-5π/6+2πn≤t≤-π/6+2πn, n∈Z.

4) sin t>√3/2;
arcsin √3/2+2πn<t<π-arcsin √3/2+2πn, n∈Z;
π/3+2πn≤t≤π-π/3+2πn, n∈Z;
π/3+2πn≤t≤2π/3+2πn, n∈Z.

5) sin t≤2/3;
-π-arcsin 2/3+2πn≤t≤arcsin 2/3+2πn, n∈Z.

6) sin t≥√2/2;
arcsin √2/2+2πn≤t≤π-arcsin √2/2+2πn, n∈Z;
π/4+2πn≤t≤π-π/4+2πn, n∈Z;
π/4+2πn≤t≤3π/4+2πn, n∈Z.

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