已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π

已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π
已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π<2x<3π/2,求（1）tanx的值
（2）（sin2x-2cos^2x）/（1-tanx）

已知sin(x+π/3)cos(x-π/3)+cos(x-π/3)sin(x-π/3)=-(2根号2/3)且π
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.