sin(x+π/6) (sin2x+sin^2x)/(1-cos2x)=sin(x+π/6)(2sinxcosx+sin^2x)/2sin^2x=[1/2 +cosx/sinx]sin(x+π/6)=[1/2+1/tanx]sin(x+π/6)=0很高兴为你解决问题
0
tanx=-2,π/2sinx=2/根号5,cosx=-1/根号5；sin2x=-4/5;sin2x+sin^2x=0sin(x+π/6)(sin2x+sin^2x)/(1-cos2x)=0
