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

2 Answers

Hence sin2x=2sinx*cosxsin2x=2sinxcosx we have that

sinx+sin2x=0=>sinx+2sinxcosx=0=>sinx(1+2cosx)=0=> sinx=0 or cosx=-1/2sinx+sin2x=0sinx+2sinxcosx=0sinx(1+2cosx)=0sinx=0orcosx=12

From sinx=0=>x=k*pisinx=0x=kπ , k \in NkN and

from cosx=-1/2=>cosx=cos(2pi/3)=>x=2*k*pi+-(2*pi)/3cosx=12cosx=cos(2π3)x=2kπ±2π3

Oct 14, 2015

x=kpi vv x=(2pi)/3+2mpi vv x=(4pi)/3+2npix=kπx=2π3+2mπx=4π3+2nπ
k,m,n in Zk,m,nZ

Explanation:

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

Using trigonometric identity:
sin2x=2sinxcosxsin2x=2sinxcosx

our equation becomes:

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

sinx=0 vv 1+2cosx=0sinx=01+2cosx=0

sinx=0 vv cosx=-1/2sinx=0cosx=12

x=kpi vv x=(2pi)/3+2mpi vv x=(4pi)/3+2npix=kπx=2π3+2mπx=4π3+2nπ
k,m,n in Zk,m,nZ