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andriymaystruk
@andriymaystruk
July 2022
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Помогите!!!!!!!!!!!!!!!!!!!!!!!!!!!
Очень, очень срочно!!!!!!!!
Лайк и т.п. !!!!!!!!
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sedinalana
Verified answer
1
sin(x-π/3)=-1
x-π/3=-π/2+2πn
x=-π/6+2πn,n∈z
2
tg(π/6-x)=-1/√3
tg(x-π/6)=1/√3
x-π/6=π/6+πn
x=π/3+πn,n∈z
3
cosx=a
2a²+5a+2=0
D=25-16=9
a1=(-5-3)/4=-2⇒cosx=-2<-1 нет решения
a2=(-5+3)/4=-1/2⇒cosx=-1/2⇒x=+-2π/3+2πn,n∈z
4
2cos2xcosx=0 U sinx≠-1⇒x≠-π/2+2πn,n∈z
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈z
cosx=0⇒x=π/2+πk,k∈z с учетом ОДЗ x=π/2+2πk,k∈z
5
sinx>-√3/2
x∈(-π/3+2πn;4π/3+2πn,n∈z)
6
tg(π/3-x)≤1/√3
tg(x-π/3)≥-1/√3
-π/6+πn≤x-π/3<π/2+πn
π/6+πn≤x<5π/6+πn
x∈[π/6+πn;5π/6+πn,n∈z)
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Answers & Comments
Verified answer
1sin(x-π/3)=-1
x-π/3=-π/2+2πn
x=-π/6+2πn,n∈z
2
tg(π/6-x)=-1/√3
tg(x-π/6)=1/√3
x-π/6=π/6+πn
x=π/3+πn,n∈z
3
cosx=a
2a²+5a+2=0
D=25-16=9
a1=(-5-3)/4=-2⇒cosx=-2<-1 нет решения
a2=(-5+3)/4=-1/2⇒cosx=-1/2⇒x=+-2π/3+2πn,n∈z
4
2cos2xcosx=0 U sinx≠-1⇒x≠-π/2+2πn,n∈z
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈z
cosx=0⇒x=π/2+πk,k∈z с учетом ОДЗ x=π/2+2πk,k∈z
5
sinx>-√3/2
x∈(-π/3+2πn;4π/3+2πn,n∈z)
6
tg(π/3-x)≤1/√3
tg(x-π/3)≥-1/√3
-π/6+πn≤x-π/3<π/2+πn
π/6+πn≤x<5π/6+πn
x∈[π/6+πn;5π/6+πn,n∈z)