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felesi
@felesi
August 2022
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Ребят, помогите, пожалуйста!
Тема: "Простейшие тригонометрические уравнения и неравенства"
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sedinalana
Verified answer
1
а)tgx=-1/√3
x=-π/6+πn,n∈z
б)tg(x-π/6)*(sin2x+1)=0
[tg(x-π/6)=0⇒x-π/6=πn⇒x=π/6+πn,n∈z
[sin2x+1=0⇒sin2x=-1⇒2x=-π/2+2πn⇒x=-π/4+πn,n∈z
в)ctgt=√3
t=π/6+πn,n∈z
2
1/2*tg2x+1/2=0
1/2*tg2x=-1/2
tg2x=-1
2x=-π/4+πn
x=-π/8+πn/2,n∈z
-7π/4≤-π/8+πn/2≤π/2
-14≤-1+4n≤4
-13≤4n≤5
-13/4≤n≤5/4
n=-3⇒x=-π/8-3π/2=-13π/8
n=-2π⇒x=-π/8-π=-9π/8
n=-1⇒x=-π/8-π/2=-5π/8
n=0⇒x=-π/8
3
√3tg(2x+π/4)≥3
tg(2x+π/4≥√3
π/3+πn≤2x+π/4<π/2+πn
π/12+πn≤2x<π/4+πn
π/24+πn/2≤x<π/8+πn/2
x∈[π/24+πn/2;π/8+πn/2,n∈z)
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Answers & Comments
Verified answer
1а)tgx=-1/√3
x=-π/6+πn,n∈z
б)tg(x-π/6)*(sin2x+1)=0
[tg(x-π/6)=0⇒x-π/6=πn⇒x=π/6+πn,n∈z
[sin2x+1=0⇒sin2x=-1⇒2x=-π/2+2πn⇒x=-π/4+πn,n∈z
в)ctgt=√3
t=π/6+πn,n∈z
2
1/2*tg2x+1/2=0
1/2*tg2x=-1/2
tg2x=-1
2x=-π/4+πn
x=-π/8+πn/2,n∈z
-7π/4≤-π/8+πn/2≤π/2
-14≤-1+4n≤4
-13≤4n≤5
-13/4≤n≤5/4
n=-3⇒x=-π/8-3π/2=-13π/8
n=-2π⇒x=-π/8-π=-9π/8
n=-1⇒x=-π/8-π/2=-5π/8
n=0⇒x=-π/8
3
√3tg(2x+π/4)≥3
tg(2x+π/4≥√3
π/3+πn≤2x+π/4<π/2+πn
π/12+πn≤2x<π/4+πn
π/24+πn/2≤x<π/8+πn/2
x∈[π/24+πn/2;π/8+πn/2,n∈z)