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

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Answer 1 · 2015-10-14T21:16:16

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 · 2015-10-14T21:19:58

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

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#