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bigfanatka
@bigfanatka
July 2022
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sin2x = sin((Pi/2)+x)
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ahta
Применим формулу sin2x=2sinxcosx, и формулу приведения sin(π/2+x)=cosx
2sinxcosx=cosx
2sinxcosx-cosx=0
cosx(2sinx-1)=0 2sinx-1=0
cosx=0, x1=π/2+πn, n∈Z sinx=0.5
x=((-1)^n)arcsin0.5+πn, n∈Z
x2=((-1)^n)(π/6+πn, n∈Z
Ответ:
n∈Z
n∈Z
4 votes
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Answers & Comments
2sinxcosx=cosx
2sinxcosx-cosx=0
cosx(2sinx-1)=0 2sinx-1=0
cosx=0, x1=π/2+πn, n∈Z sinx=0.5
x=((-1)^n)arcsin0.5+πn, n∈Z
x2=((-1)^n)(π/6+πn, n∈Z
Ответ: n∈Z
n∈Z