1) Sin(2x-pi/6)<1/2
2)cos(pi/3+1/2x)>= - корень из 3/2
3)cos(3x-pi/4)<= - корень из 2/2
4)корень из 2 sin(x/2+pi/4)>=1
Ответы
Ответ дал:
1
1
Sin(2x-pi/6)<1/2
5π/6+2πk<2x-π/6<13π/6+2πk
π+2πk<2x<4π/6+2πk
π/2=πk<x<7π/12+πk,k∈z
2
cos(pi/3+1/2x)≥-√ 3/2
7π/6+2πk≤π/3+x/2≤17π/6+2πk
5π/6+2πk≤x/2≤5π/2+2πk
5π/3+4πk≤x≤5π+4πk,k∈z
3
cos(3x-pi/4)≤-√ 2/2
3π/4+2πk≤3x-π/4≤5π/4+2πk
π+2πk≤3x≤3π/2+2πk
π/3+2πk/3≤x≤π/2+2πk/3,k∈z
4
√ 2 sin(x/2+pi/4)≥1
sin(x/2+π/4)≥1/2
π/6+2πk≤x/2+π/4≤5π/6+2πk
-π/12+2πk≤x/2≤7π/12+2πk
-π/6+4πk≤x≤7π/6+4πk,k∈z
Sin(2x-pi/6)<1/2
5π/6+2πk<2x-π/6<13π/6+2πk
π+2πk<2x<4π/6+2πk
π/2=πk<x<7π/12+πk,k∈z
2
cos(pi/3+1/2x)≥-√ 3/2
7π/6+2πk≤π/3+x/2≤17π/6+2πk
5π/6+2πk≤x/2≤5π/2+2πk
5π/3+4πk≤x≤5π+4πk,k∈z
3
cos(3x-pi/4)≤-√ 2/2
3π/4+2πk≤3x-π/4≤5π/4+2πk
π+2πk≤3x≤3π/2+2πk
π/3+2πk/3≤x≤π/2+2πk/3,k∈z
4
√ 2 sin(x/2+pi/4)≥1
sin(x/2+π/4)≥1/2
π/6+2πk≤x/2+π/4≤5π/6+2πk
-π/12+2πk≤x/2≤7π/12+2πk
-π/6+4πk≤x≤7π/6+4πk,k∈z
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