已知函数f(x)=(√3sinωx+cosωx)cosωx+1/2(ω>0)的最小正周期为π,则函数f(x)的单调递增区间为?

题目

已知函数f(x)=(√3sinωx+cosωx)cosωx+1/2(ω>0)的最小正周期为π,则函数f(x)的单调递增区间为?

答案

f(x）=根号3/2*sinwx +1/2*coswx +1
= sin(wx+π/6)+1
T=2π/w =π , 所以, w=2
f(x)= sin(2x+π/6)+1
-π/2 +2kπ <= 2x +π/6 <=π/2 +2kπ
解得, -π/3 +kπ <=x<= π/6 +kπ, k∈Z
递增区间为 [ -π/3 +kπ , π/6 +kπ], k∈Z