已知cos（π/4+x）=3/5，且7π/12<x<7π/4，求（2sin^x+sin2x）/1-tanx.

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17π/12＜x＜7π/4，得5π/3＜x+π/4＜2π
cos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-cos²(x+π/4)]=-√[1-(3/5)²]=-4/5
sin(2x)=-cos(2x+π/2)=-cos[2(x+π/4)]=1-2cos²(x+π/4)=1-2•(3/5)²=7/25
[sin(2x)+2sin²x]/(1-tanx)
=2(sinxcosx+sin²察埋x)/(1-sinx/cosx)
=2(cosx+sinx)/(1/sinx-1/cosx)
=2(cosx+sinx)sinxcosx/败游蚂(cosx-sinx)
=cos(x-π/4)sin(2x)/cos(x+π/4)
=-4/5•7/25/磨郑(3/5)
=-28/75

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