已知函数f(x)=(sinωx)^2+√3sinωx·sin(ωx+π/2)(ω>0)的最小正周期为π 求f(x)的单调增区间

Question

如题所述

Answer 1

1. f(x)=sin²ωx+根号3sinωxsin[ωx+π/2]
=1/2-(1/2)cos2wx+√3sinwxcoswx
=(√3/2)sin2wx-(1/2)cos2wx+1/2
=sin(2wx-π/6)+1/2
(1) 最小正周期T=2π/2w=π w=1
(2) x∈[0,2π/3] 2x-π/6∈[-π/6, 7π/6]
sin(2x-π/6)∈[-1/2, 1]
f(x)的值域为[0, 3/2]追问

貌似第二问你看错题目了亲，是f(x)的单调增区间，麻烦重写一下，谢谢！

Answer 2

已知函数f(x)=sin²ωx)+(√3)sinωx·sin(ωx+π/2)(ω>0)的最小正周期为π 求f(x)的单调增区间
解：f(x)=(1-cos2ωx)/2+(√3)sinωxcosωx=(1/2)+[(√3/2)sin2ωx-(1/2)cos2ωx]
=(1/2)+[sin2ωxcos(π/6)-cos2ωxsin(π/6)]=(1/2)+sin(2ωx-π/6)
T=2π/2ω=π/ω=π，故ω=1，即有f(x)=1/2+sin(2x-π/6)
单增区间：由-π/2+2kπ≦2x-π/6≦π/2+2kπ，得-π/3+2kπ≦2x≦2π/3+2kπ，
故单增区间为：-π/6+kπ≦x≦π/3+kπ，k∈Z