Question

三角函数，求过程

Answer 1

1 定义域 cos(2x-π/3)>0
2kπ-π/2< 2x-π/3<2kπ+π/2,k∈Z
∴定义域为{x|kπ-π/12<x<kπ+5π/12,k∈Z}

2 当kπ-π/12<x≤kπ+π/6时，
cos(2x-π/3)递增，∴lgcos(2x-π/3)递增
∴f(x)递增区间为（kπ-π/12,kπ+π/6]
f(x)递减区间为[kπ+π/6,kπ+5π/12),k∈Z

3 x=π/6 和x=-π/6不对应 一个存在一个不存在
∴f(x)非奇偶函数

4f(x+π)=lgcos[2(x+π)-π/3]
=lgcos(2x+2π-π/3)
=lgcos(2x-π/3)
=f(x)
∴f(x)是周期函数，周期为π