若cos（π⼀4+x)=3⼀5,17⼀12π<x<7⼀4π，求sin2x+2cosx^2的值

解：
∵(17/12)π∴(5/3)π∴sin(x+π/4)<0
∵cos(x+π/4)=3/5
∴sin(x+π/4)=-4/5
∴tan(x+π/4)=-4/3
∴[tanx+tan(π/4)]/[1-tanxtan(π/4)]=-4/3
∴(1+tanx)/(1-tanx)=-4/3
∴3(tanx+1)=4(tanx-1)
∴3tanx+3=4tanx-4
∴tanx=7
∴sin(2x)+2(cosx)^2
=[sin(2x)+2(cosx)^2]/1
=[2sinxcosx+2(cosx)^2]/[(sinx)^2+(cosx)^2]
=(2tanx+2)/[(tanx)^2+1]
=(14+2)/(49+1)
=8/25.

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-sin²(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+sin²x)/(1-sinx/cosx)
=2(cosx+sinx)/(1/sinx-1/cosx)
=2(cosx+sinx)sinxcosx/(cosx-sinx)
=cos(x-π/4)sin(2x)/cos(x+π/4)
=-4/5•7/25/(3/5)
=-28/75