解
f(x)=2√3cos²x-2sinxcosx-√3
=√3(2cos²x-1)-sin2x
=√3cos2x-sin2x
=2(√3/2cos2x-1/2sin2x)
=2(cos2xcosπ/6-sinπ/6sin2x)
=2cos(2x+π/6)
∴T=2π/2=π是f(x)的最小正周期
∵/cos(2x+π/6)/≤1
∴f(x)的最小值为：-2
当-π+2kπ≤2x+π/6≤0+2kπ
即-7π/12+kπ≤x≤-π/12+kπ时
f(x)是增函数
∴增区间为[kπ-7π/12,kπ-π/12](k∈z)