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locer
@locer
August 2022
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РЕШИТЕ УРАВНЕНИЯ , пожалуйста
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m11m
А) 2cos²x +2cosx+sin²x=0
2cos²x +2cosx+1-cos²x=0
cos²x+2cosx+1=0
(cosx+1)²=0
cosx+1=0
cosx=-1
x=π+2πn
б) sin3xcosx + sinxcos3x=0
sin(3x+x)=0
sin4x=0
4x=πn
x=
π
n
4
в) cos2x cosx-sin2x sinx=-1
cos(2x+x)=-1
cos3x=-1
3x=π+2πn
x=
π
+
2π
n
3 3
1 votes
Thanks 0
locer
http://znanija.com/task/11607391 помоги еще с этим,пожалуйста:))
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Answers & Comments
2cos²x +2cosx+1-cos²x=0
cos²x+2cosx+1=0
(cosx+1)²=0
cosx+1=0
cosx=-1
x=π+2πn
б) sin3xcosx + sinxcos3x=0
sin(3x+x)=0
sin4x=0
4x=πn
x=πn
4
в) cos2x cosx-sin2x sinx=-1
cos(2x+x)=-1
cos3x=-1
3x=π+2πn
x=π + 2πn
3 3