We know that sin2x=2sinxcosx
(search the net for proof if you wish)
So the original equation becomes
2sinxcosx-sinx=0
The two terms both have sinx that can be taken out to get:
sinx(2cosx-1)=0
This is true if sinx=0 or 2cosx-1=0 , rewritten: cosx=1/2
sinx=0 than x=2kπ
cosx=1/2 than x=π/3+2kπ
where k is an integer
