cos(π/4+x)=3/5,
∴sin2x=-cos[2(π/4+x)]=-[2*(3/5)^2-1]=7/25,
17π/12∴5π/3<π/4+x<2π,
∴sin(π/4+x)=-4/5.
[sin2x+2(sinx)^2]/(1-tanx)
=[2sinxcosx+2(sinx)^2]/(1-sinx/cosx)
=2sinxcosx(cosx+sinx)/(cosx-sinx)
=sin2xsin(π/4+x)/cos(π/4+x)
=(7/25)*(-4/5)/(3/5)
=-28/75.