What are the roots of $\sin(ax) + \sin((a + 2)x)$? I was playing around with $\sin(5x) + \sin(7x)$, wondering where the roots of the function are. I graphed it on wolframalpha and from the list of solutions I guessed that the solutions to $\sin(5x) + \sin(7x) = 0$ are exactly the roots of $\sin(6x)$.
After playing around with similar values, I have to wonder, is it true that solving 
$$\sin(a\cdot x) + \sin((a + 2)\cdot x) = 0$$
is equivalent to
$$\sin((a + 1)\cdot x) = 0$$

This problem seems to want induction.
It is true that the roots of $\sin(x) + \sin(3x)$ are the same roots as $\sin(2x)$ (again checked via wolframalpha). So that takes care of the base case.
However, I have no idea how to get from the $\sin(a\cdot x) + \sin((a+2)\cdot x)$ to the $\sin((a+1)\cdot x)$ for any $a$. It's just been guesswork.

Is it true that $\sin(a\cdot x) + \sin((a + 2)\cdot x)$ has the same roots as $\sin((a + 1)\cdot x)$?

There is really nothing special about the pair $(a,a+2)$, so we look at $\sin(ax)+\sin(bx)=0$, or equivalently $\sin(ax)=\sin(-bx)$.
When are the sines of two numbers equal? We have $\sin s=\sin t$ if $t$ is of the form $s+2k\pi$, or if $t$ is of the form $\pi-s+2k\pi$, with $k$ an arbitrary integer.
Now we can write down the general solution of $\sin(ax)=\sin(-bx)$. 
