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Fredyson
@Fredyson
November 2021
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Вот последняя часть.Помогите)
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sedinalana
Verified answer
3
1)6x=+-π/6+2πn
x=+-π/36+πn/3,n∈z
2)x/3+π/6=-π/2+2πn
x/3=-2π/3+2πn
x=-2π+6πn,n∈z
3)3sinx+√3cosx=0/cosx
3tgx+√3=0
tgx=-√3/3
x=-π/6+πn,n∈z
4)4(1-cos²x)-11cosx-1=0
4-4cos²x-11cosx-1=0
cosx=a
4a²+11a-3=0
D=121+48=169
a1=(-11-13)/8=-3⇒cosx=-3<-1 нет решения
a2=(-11+13)/8=1/4⇒cosx=1/4⇒x=+-arccos0,25+2⇒n,n∈z
5)2sin2xcosx+sinx=0
4sinxcos²x+sinx=0
sinx*(4cos²x+1)=0
sinx=0⇒x=πn,n∈z
4cos²x+1=0⇒cos²x=-1/4 нет решения
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Answers & Comments
Verified answer
31)6x=+-π/6+2πn
x=+-π/36+πn/3,n∈z
2)x/3+π/6=-π/2+2πn
x/3=-2π/3+2πn
x=-2π+6πn,n∈z
3)3sinx+√3cosx=0/cosx
3tgx+√3=0
tgx=-√3/3
x=-π/6+πn,n∈z
4)4(1-cos²x)-11cosx-1=0
4-4cos²x-11cosx-1=0
cosx=a
4a²+11a-3=0
D=121+48=169
a1=(-11-13)/8=-3⇒cosx=-3<-1 нет решения
a2=(-11+13)/8=1/4⇒cosx=1/4⇒x=+-arccos0,25+2⇒n,n∈z
5)2sin2xcosx+sinx=0
4sinxcos²x+sinx=0
sinx*(4cos²x+1)=0
sinx=0⇒x=πn,n∈z
4cos²x+1=0⇒cos²x=-1/4 нет решения