Question #52d88

We know that :

`sina+sinb=2sin((a+b)/2)cos((a-b)/2)`

so

`sin3x+sinx=0 iff 2sin((3x+x)/2)cos((3x-x)/2)=0 iff`

`2sin2xcosx=0iff sin2xcosx=0 iff`

`sin2x=0 or cosx=0`

Let's start with `sin2x=0`

`sin2x=0 iff 2x=2kpi or 2x=2kpi+piiff 2x=kpi iff`

`x=(kpi)/2, kin ZZ`

because `x in[0,pi],` `k` can be `0 or 1 or 2`

so `x` can be `0 or pi/2 or pi`

Let's start with `cosx=0` now :

`cosx=0 iff cosx=cos(pi/2) iff x=2kpi+-pi/2` the only accepted solution from here is `x=pi/2`

Now we gather all of our solutions so we have :

`x=0 or x=pi/2 or x=pi`

A different approach.

Using de Moivre's identity

`sin x=(e^(ix)-e^(-ix))/(2i)` we have

`(e^(3ix)-e^(-3ix))/(2i)+(e^(ix)-e^(-ix))/(2i)=0`

and now using `a^3-b^3=(a-b)(a^2+ab+b^2)` we have

`(e^(ix)-e^(-ix))(e^(2ix)+1+e^(-2ix))+(e^(ix)-e^(-ix))=0` or

`(e^(ix)-e^(-ix))(e^(2ix)+2+e^(-2ix))=0`

then

`{(e^(ix)-e^(-ix)=0),(e^(2ix)+2+e^(-2ix)=0):}`

and from

`e^(ix)-e^(-ix)=0->e^(-ix)(e^(2ix)-1)=0->2x=0+2kpi->x = kpi` because `e^(-ix) ne 0`

and from

`e^(2ix)+2+e^(-2ix)=0` calling `y = e^(2ix)` we have

`1/y(y+1)(y+1)=0->y = e^(2ix)=-1` because `e^(-2ix) ne 0`

so

`e^(2ix)=-1->2x=pi+2kpi->x = pi/2+kpi`

Resuming, the solutions are:

`x=(kpi)uu(pi/2+kpi),k=0,pm1,pm2,cdots`

and in the interval `[0,pi]` we have the set

`x = {0,pi/2,pi}`