sin(3x) cancels:
sin(5x)-sin(x)-sin(7x)=0,
sin(5x)=sin(x)+sin(7x).
sin(A+B)=sinAcosB+cosAsinB;
sin(A-B)=sinAcosB-cosAsinB;
sin(A+B)-sin(A-B)=2cosAsinB.
Let A+B=7x and A-B=5x, then 2A=12x, so A=6x and 2B=2x, so B=x.
sin(7x)-sin(5x)=2cos(6x)sin(x).
Therefore 0=sin(x)+2cos(6x)sin(x), 0=sin(x)(1+2cos(6x)).
2cos(6x)=-1, cos(6x)=-½, 6x=(π-π/3)+2πn=2π/3+2πn, x=(2π/3+2πn)/6=π/9+nπ/3, where n is an integer.
Also 6x=π+π/3+2πn=4π/3+2πn, x=(4π/3+2πn)/6=2π/9+nπ/3.
We are not given any limits for x, so we have to cover all possible solutions:
x=π/9+nπ/3, 2π/9+nπ/3.