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Zhumabekanele
@Zhumabekanele
August 2022
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Sinx + sin2x + sin3x = 0; sinx + sin3x = 2sin2x; sin3x - 2sinx = 0; можете решить.... пожалуйста
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WRTLN
1) sinx+sin3x+sin2x=0
2*sin2x*cosx+sin2x=0
sin2x(2cosx+1)=0
sin2x=0
2x=Pi*n, n
∈Z
x=Pi*n/2, n
∈Z
2cosx+1=0
cosx=-1/2
x=±2Pi/3+2Pi*n, n
∈Z
2)sinx+sin3x=2sin2x
2sin2x*cosx-2sin2x=0
2sin2x(cosx-1)=0
2sin2x=0
sin2x=0
2x=Pi*n, n∈Z
x=Pi*n/2, n∈Z
cosx-1=0
cosx=1
x=2Pi*n, n∈Z
3)sin3x-2sinx=0
Это уравнение пока до меня не Дошло.
1 votes
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Answers & Comments
2*sin2x*cosx+sin2x=0
sin2x(2cosx+1)=0
sin2x=0
2x=Pi*n, n∈Z
x=Pi*n/2, n∈Z
2cosx+1=0
cosx=-1/2
x=±2Pi/3+2Pi*n, n∈Z
2)sinx+sin3x=2sin2x
2sin2x*cosx-2sin2x=0
2sin2x(cosx-1)=0
2sin2x=0
sin2x=0
2x=Pi*n, n∈Z
x=Pi*n/2, n∈Z
cosx-1=0
cosx=1
x=2Pi*n, n∈Z
3)sin3x-2sinx=0
Это уравнение пока до меня не Дошло.