METHOD 1
sin(5x)-sin(3x)+sin(x)=0.
sin(A+B)=sinAcosB+cosAsinB and sin(A-B)=sinAcosB-cosAsinB.
sin(A+B)-sin(A-B)=2cosAsinB.
Let A+B=5x and A-B=3x, then 2A=8x, so A=4x and 2B=2x, B=x.
Therefore:
sin(5x)-sin(3x)+sin(x)=0=2cos(4x)sin(x)+sin(x)=sin(x)(2cos(4x)+1).
Therefore sin(x)=0, so x=0, π.
And cos(4x)=-½, 4x=π-π/3+2nπ=2π/3+2nπ, x=π/6+nπ/2; also 4x=π+π/3+2nπ, x=π/3+nπ/2, where n is an integer: 0≤n≤3. x=π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3, 11π/6.
Total solution, x=0, π/6, π/3, 2π/3, 5π/6, π, 7π/6, 4π/3, 5π/3, 11π/6.
METHOD 2
sin(5x)=sin(4x+x)=sin(4x)cos(x)+cos(4x)sin(x)=
2sin(2x)cos(2x)cos(x)+(1-2sin2(2x))sin(x)=
4sin(x)cos2(x)(1-2sin2(x))+(1-8sin2(x)cos2(x))sin(x)=
4sin(x)(1-sin2(x))(1-2sin2(x))+(1-8sin2(x)cos2(x))sin(x). Note that sin(x) is a common factor:
sin(x)[4(1-sin2(x))(1-2sin2(x))+1-8sin2(x)(1-sin2(x))].
sin(3x)=sin(2x)cos(x)+cos(2x)sin(x)=2sin(x)(1-sin2(x))+(1-2sin2(x))sin(x)=
sin(x)[2(1-sin2(x))+1-2sin2(x)].
We can divide through by sin(x), remembering that sin(x)=0 is a solution and we can write the original equation:
sin(5x)/sin(x)+1=sin(3x)/sin(x), therefore:
4(1-sin2(x))(1-2sin2(x))+1-8sin2(x)(1-sin2(x))+1=2(1-sin2(x))+1-2sin2(x).
4(1-3sin2(x)+2sin4(x))+1-8sin2(x)+8sin4(x)+1=2-2sin2(x)+1-2sin2(x),
6-20sin2(x)+16sin4(x)=3-4sin2(x),
3-16sin2(x)+16sin4(x)=0=(4sin2(x)-3)(4sin2(x)-1).
Therefore sin2(x)=¾, ¼; sin(x)=±√3/2,±½; x=±π/3, ±π/6, that is, π/6, π/3, 2π/3, 5π/6, 7π/6, 4π/3, 5π/3, 11π/6.
And finally we have to remember sin(x)=0 is also a solution, so in addition we have x=0, π.