是sin(x+π/3)cos(x-π/3)+cos(x+π/3)sin(x-π/3)=-(2√2/3)吧？如果是，则上式可化为sin2x=-2√2/3，∵π<2x<3π/2，∴cos2x=-1/3，(1) tanx=sin2x/(1+cos2x)=-√2。(2) 2cos²x=1+cos2x=2/3，∴原式=(-2√2/3-2/3)/(1-√2)=(6+4√2)/3。
