Question

How do you solve the identity `sin(x) + sin (2x) = 0`?

Answer 1

Hence `sin2x=2sinx*cosx` we have that

#sinx+sin2x=0=>sinx+2sinxcosx=0=>sinx(1+2cosx)=0=> sinx=0 or cosx=-1/2#

From `sinx=0=>x=k*pi` , `k \in N` and

from `cosx=-1/2=>cosx=cos(2pi/3)=>x=2*k*pi+-(2*pi)/3`

Answer 2

`x=kpi vv x=(2pi)/3+2mpi vv x=(4pi)/3+2npi`
`k,m,n in Z`

Explanation:

First of all, this is not identity, it's an equation.

Using trigonometric identity:
`sin2x=2sinxcosx`

our equation becomes:

`sinx+2sinxcosx=0`
`sinx(1+2cosx)=0`

`sinx=0 vv 1+2cosx=0`

`sinx=0 vv cosx=-1/2`

`x=kpi vv x=(2pi)/3+2mpi vv x=(4pi)/3+2npi`
`k,m,n in Z`