若cos(π/4+x)=3/5 17π/12

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 再问： 不对吧 是cos²x 不是sin²x 而且是cos(π/4+x）不是cos(x-π/4)你再看看 别玩复制粘贴 OK？ 再答： cos(π/4+x) =cosπ/4cosx-sinπ/4sinx =√2/2(cosx-sinx) =3/5 cosx-sinx=3√2/5 …………（1） （cosx-sinx）^2=(cosx)^2+(sinx)^2-2sinxcosx=1-2sinxcosx=18/25 2sinxcosx=1-18/25=7/25 所以sin2x=2sinxcosx=7/25 由（1）得 sinx=cosx-3√2/5 两边平方 sin²x=1-cos²x=cos²x-6√2/5*cosx+18/25 2cos²x-6√2/5*cosx-7/25=0 50cos²x-30√2cosx-7=0 cosx=(3√2±4√2)/10 cosx=-√2/10或7√2/10 17π/12