Question

已知函数f(x)=2cos^2ωx+2√3sinωx cosωx+3(ω＞0)的最小正周期为π （1）求ω的值

（2）求函数f(x)的单调区间、、、

Answer 1

f(x)=2cos^2ωx+2√3sinωx
cosωx+3
=cos2ωx+√3sin2ωx
+4
=2sin(2ωx+π/6)+4
(1)因为
T=2π/2ω=π
所以
ω=1
(2)
f(x)=2sin(2x+π/6)+4
单调增为2kπ-π/2≤2x+π/6≤2kπ+π/2
2kπ-4π/6≤2x≤2kπ+2π/6
kπ-π/3≤x≤kπ+π/6
(k∈Z)
单调减为2kπ+π/2≤2x+π/6≤2kπ+3π/2
2kπ+2π/6≤2x≤2kπ+8π/6
kπ+π/6≤x≤kπ+2π/3
(k∈Z)
故
函数f(x)的单调递增区间：[kπ-π/3，kπ+π/6]
(k∈Z)
同理可得：函数f(x)的单调递减区间：[kπ+π/6，kπ+2π/3]
(k∈Z)