解 f(x)=√3cos²x+sinxcosx-√3/2
=√3*(1+cos2x)/2+(1/2)sin2x-√3/2
=(1/2)sin2x+(√3/2)cos2x
=sin(2x+π/3)
∴T=π
單增區(qū)間：-π/2+2kπ≤2x+π/3≤π/2+2kπ,k∈Z
-5π/6+2kπ≤2x≤π/6+2kπ,k∈Z
-5π/12+kπ≤x≤π/12+2kπ,k∈Z
即為：[-π/12+kπ,5π/12+kπ],k∈Z
(2)結(jié)合圖像,5π/12+17π/12+29π/12=51π/12