3sinθtanθ=8,（0＜θ＜π）,则cos（θ-π/4）=

∵0＜θ＜π
∴sinθ>0
∵3sinθtanθ=8
==>3(sinθ)^2/cosθ=8
==>(sinθ)^2=8cosθ/3
==>(cosθ)^2+8cosθ/3=1 (∵(sinθ)^2+(cosθ)^2=1)
==>3(cosθ)^2+8cosθ-3=0
==>(3cosθ-1)(cosθ+3)=0
==>3cosθ-1=0 (∵cosθ+3>0)
∴cosθ=1/3,sinθ=√(8cosθ/3)=2√2/3
故cos(θ-π/4)=cosθcos(π/4)+sinθsin(π/4) (应用差角公式)
=(1/3)(√2/2)+(2√2/3)(√2/2)
=(√2+4)/6.