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LOLgreen
@LOLgreen
July 2022
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sedinalana
3
f`(x)=2sinx*cosx-1
sin2x-1=0
sin2x=1
2x=π/2+2πk
x=π/4+πk,k∈z
4
y(2π)=-sin2π+√3=√3
y`=-cosx
y`(2π)=-cos2π=-1
y=√3-1(x-2π)=-x+√3+2π
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Answers & Comments
f`(x)=2sinx*cosx-1
sin2x-1=0
sin2x=1
2x=π/2+2πk
x=π/4+πk,k∈z
4
y(2π)=-sin2π+√3=√3
y`=-cosx
y`(2π)=-cos2π=-1
y=√3-1(x-2π)=-x+√3+2π