sin5x+sinx=sin3x

Question

Find all the solutions in the internal [0,2pie)

Answer 1

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.