Question

已知函数f(x)=√3sinωx+cosωx(ω>0)的最小正周期为π,则函数f(x)的单调递增区间为

已知函数f(x)=√3sinωx+cosωx(ω>0)的最小正周期为π,则函数f(x)的单调递增区间为

Answer 1

f(x)=2sin(wx+π/6)
T=2π/w=π
则：w=2
所以，f(x)=2sin(2x+π/6)
递增区间
-π/2+2kπ<2x+π/6<π/2+2kπ
-2π/3+2kπ<2x<π/3+2kπ
-π/3+kπ<x<π/6+kπ
所以，递增区间为（-π/3+kπ，π/6+kπ）k∈Z

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Answer 2

f(x)=2sin(wx+π/6)
T=2π/w=π
则：w=2
所以，f(x)=2sin(2x+π/6)
递增区间
-π/2+2kπ<2x+π/6<π/2+2kπ
-2π/3+2kπ<2x<π/3+2kπ
-π/3+kπ<x<π/6+kπ
所以，递增区间为（-π/3+kπ，π/6+kπ）k∈Z

Answer 3

T=2兀/w w=2 f(x)=2sin(2x﹢兀/6)