2sinx+3sin(2y+x)=0,求5tan(x+y)+tany=

x=（x+y）-y ， 2y+x=（x+y）+y
2sinx+3sin(2y+x)=0 =>2sin[（x+y）-y] +3sin[（x+y）+y]=0
展开：2sin（x+y）cosy-2cos（x+y）siny+3sin（x+y）cosy+3cos（x+y）siny=0
整理得：5sin（x+y）cosy+cos（x+y）siny=0
两边同时除以cos（x+y）cosy
得5tan(x+y)+tany=0

2sinx+3sin(2y+x)
=2sin[(x+y-y)]+3sin[(x+y)+y]
=2sin(x+y)cosy-2cos(x+y)siny+3sin(x+y)cosy+3cos(x+y)siny
=5sin(x+y)cosy+cos(x+y)siny
=cos(x+y)cosy[5tan(x+y)+tany]

条件给少了，根据化简，应该是要求给出的x、y值能使得cos(x+y)cosy≠0，至少还要y≠kπ+π/2、x+y≠kπ+π/2什么的

这样由2sinx+3sin(2y+x)=0就有cos(x+y)cosy[5tan(x+y)+tany]=0
然后cos(x+y)cosy≠0，所以：5tan(x+y)+tany=0

2sinx+3sin(2y+x)=0
2sinx+3sinxcos2y+3cosxsin2y=0
sinx(2+3cos2y)+3cosxsin2y=0
sinx(2+2cos2y+cos2y)+3cosxsin2y=0
sinx(4(cosy)^2+cos2y)+3cosxsin2y=0
sinx(5(cosy)^2-(siny)^2)+3cosxsin2y=0
5sinx(cosy)^2-sinx(siny)^2+6cosxsinycosy=0
5sinx(cosy)^2+5cosxsinycosy+cosxsinycosy-sinx(siny)^2=0
5sin(x+y)cosy+cos(x+y)siny=0 [同除 cos(x+y)cosy]得到