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
Abraham Lincoln, i've read that story a long time ago myself and I remember who said that.
<u>Answer:</u>
x = 4 (extraneous solution)
<u>Step-by-step explanation:</u>

This solution is extraneous. Reason being that even if it can be solved algebraically, it is still not a valid solution because if we substitute back
, we will get two fractions with zero denominator which would be undefined.