1
f(x)=cosx(cosx-根号下3sinx)-1/2
=cos²x-√3sinxcosx-1/2
=1/2(1+cos2x)-√3/2*sin2x-1/2
=1/2cos2x-√3/2*sin2x
=cos(2x+π/3)
f(π/6)=cos(2*π/6+π/3)=cos2π/3=-1/2
2
∵0∴2x+π/3=π,即x=π/3时,f(x)取得π最小值-1