设函数f(x)=sin(πx/4-π/6)-2cos^2πx/8+1.求f(x)的最小正周期
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设函数f(x)=sin(πx/4-π/6)-2cos^2πx/8+1.求f(x)的最小正周期
f(x)= sin(πx/4-π/6) - [2cos^2πx/8-1] = sin(πx/4-π/6) - cosπx/4
最小正周期应该就是2π/(π/4)=8
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