求解一道三角函数体

别人提问是需要有详细解答过程的：
sin[(x+y)+x]=5sin[(x+y)-x]
sin(x+y)·cosx＋cos(x+y)·sinx＝5·sin(x+y)·cosx－5·cos(x+y)·sinx
4·sin(x+y)·cosx＝6·cos(x+y)·sinx
两边同除以cos(x+y)·cosx得2·tan(x+y)=3·tanx

sin(2x+y)=sin[(x+y)+x]=sin(x+y)cosx+cos(x+y)sinx①
5siny=5sin(x+y-x)=5[sin(x+y)cosx-cos(x+y)sinx]②
①=②,通过移项得到4sin(x+y)cosx=6cos(x+y)sinx,
即2sin(x+y)cosx=3cos(x+y)sinx,两边同时除以cos(x+y)cosx就得到2tan(x+y)=3tanx.