f(x)=√3/2*sin2x-(1+cos2x)/2+1/4=√3/2*sin2x-1/2*cos2x-1/4=sin2x*cos(π/6)-cos2x*sin(π/6)-1/4=sin(2x-π/6)-1/4所以f(x)的最小正周期T=2π/2=π当2x-π/6=2kπ+π/2时,即x=kπ+π/3时,f(x)最大,f(x)max=1-1/4...