已知函数f(x)=sin(πx/3-6/π)-2cos^2πx/6
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已知函数f(x)=sin(πx/3-6/π)-2cos^2πx/6
求最小正周期和单调递增区间
求最小正周期和单调递增区间
f(x)=sin(πx/3-6/π)-2cos^2πx/6
=sinπx/3cosπ/6-cosπx/3sinπ/6-(1+cosπx/3)
=√3/2sinπx/3-3/2cosπx/3-1
=√3(1/2sinπx/3-√3/2cosπx/3)-1
=√3sin(πx/3-π/3)-1
f(x)最小正周期T=2π/(π/3)=6
由2kπ-π/2≤πx/3-π/3≤2kπ+π/2,k∈Z
得6k-1/2≤x≤6k+5/2,k∈Z
∴f(x)递增区间为[6k-1/2,6k+5/2],k∈Z
=sinπx/3cosπ/6-cosπx/3sinπ/6-(1+cosπx/3)
=√3/2sinπx/3-3/2cosπx/3-1
=√3(1/2sinπx/3-√3/2cosπx/3)-1
=√3sin(πx/3-π/3)-1
f(x)最小正周期T=2π/(π/3)=6
由2kπ-π/2≤πx/3-π/3≤2kπ+π/2,k∈Z
得6k-1/2≤x≤6k+5/2,k∈Z
∴f(x)递增区间为[6k-1/2,6k+5/2],k∈Z
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