由题意可知f(x1)=f(x)min=-1=>sin(π/2x1+π/3)=-1=>π/2x1+π/3=2k1π-π/2=>x1=1/(4k1-5/3)同理f(x2)=f(x)max=1=>sin(π/2x2+π/3)=1=>π/2x2+π/3=2k2π+π/2=>x2=1/(4k2+1/3)|x1-x2|=|1/(4k1-5/3)-1/(4k2+1/3)|...