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

#sin(2x)+sinx=0#

#2sinxcosx+sinx=0#

#sinx(2cosx+1)=0#

#sinx=0,qquadcosx=-1/2#

Here's a unit circle to remind us of where the sine and cosine values are:

This means that:

#x=0,pi,(2pi)/3,(4pi)/3#

Since these values are the same after any full #2pi# rotation, we write #+2pik# after every solution to represent that the answer will stay the same after any rotation added or subtracted:

#x=0+2pik,qquadpi+2pik,qquad(2pi)/3+2pik,qquad(4pi)/3+2pik#

Technically, our answers are complete, but we can go a little further.

We can see a pattern in some of the solutions: #0, (2pi)/3, (4pi)/3#

Since these are multiples of #(2pi)/3#, we can rewrite these three solutions as #(2pi)/3*k#. Now, our solutions are:

#x=(2pi)/3*k,qquadpi+2pik#

Since #0# is still a solution, there is another pattern: #0, pi, 2pi#

We can rewrite this again as multiples of #pi#, or #pik#. Now, our final-final solutions are:

#x=(2pi)/3*k,qquadpik#

These are the solutions. Hope this helped!

put #sin(2x)=2sin(x)cos(x)#

equation will now be reduced to

#2sin(x)cos(x)+sin(x)=0#

taking #sin(x)# common we get

#sin(x)(2cos(x)+1)=0#

two possibilities are there

either #sin(x)=0# or #2cos(x)+1=0#

if #sin(x)=0# then #x=npi#

if #2cos(x) +1 = 0# then #x=2npi + (2pi)/3 or 2npi -(2pi)/3 #

so solution to the equation is

#x= 2npi + (2pi)/3 or 2npi -(2pi)/3 or npi#