f(X)=2根号3sinwxcoswx-2cos^wx
=√3sin(2wx)-(1+cos2wx)
=2[(√3/2)sin(2wx)-(1/2)cos(2wx)]-1
=2sin(2wx-π/6)-1
最小正周期为 2π/|2w|=π/|w|=π
所以 w=1
f(x)=2sin(2x-π/6)-1
单增：
2x-π/6∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-π/6,kπ+π/3]
所以
单调增区间为
[kπ-π/6,kπ+π/3] k∈z