当x∈（π,2π）时,x-π∈(0,π),f(x-π)=f(x-2π)+sin(x-π)=f(x-2π)-sinx=x-π
f(x-2π)=sinx+x-π,f(x)=sin(x+2π)+(x+2π)-π=sinx+π+x
当x∈（2π,3π）时,x-π∈(π,2π),f(x-π)=f(x-2π)+sin(x-π)=f(x-2π)-sinx=sinx+π+x
f(x-2π)=2sinx+x+π,f(x)=2sin(x+2π)+(x+2π)+π=2sinx+3π+x
x=2π时,f(2π)=f(π)+sin2π,f(2π)=f(π)=f(0)
综上,x∈（π,2π）时,f(x)=sinx+π+x
x∈（2π,3π）时,f(x)=2sinx3+π+x