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 $\sin((a + 1)\cdot x)$?
 A: Note the sum to product identity: $\sin A + \sin B = 2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$. 
Hence, $\sin(ax) + \sin((a+2)x) = 2\sin((a+1)x)\cos(x)$. 
So the roots of $\sin(ax) + \sin((a+2)x)$ are the roots of $\sin((a+1)x)$ along with the roots of $\cos(x)$. 
If $x = \dfrac{\pi}{2}+\pi k$ (the roots of $\cos x$) are also roots of $\sin((a+1)x)$, then we are only left with the roots of $\sin((a+1)x)$. This happens iff $\sin\left((a+1)(\dfrac{\pi}{2}+\pi k)\right) = 0$, i.e. $a$ is odd (assuming $a \in \mathbb{Z}$). 
Here are the graphs of $\sin(2x)+\sin(4x)$ and $\sin(3x)$. Note that $\sin(2x)+\sin(4x)$ has more roots.
A: 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 $\sin(ax)=\sin(-bx)$. 
A: We transform that sum into a product. An easy way to deduce it is: We begin with $\sin X + \sin Y$. Let $X = c + d$ and $Y = c - d$. This gives us $c = \frac{X + Y}{2}$ and $d = \frac{X - Y}{2}$. This way: $$\begin{align} \sin X + \sin Y &= \sin(c+d)+ \sin(c-d) \\ &= \sin c \cos d + \sin c \cos d + \sin c \cos d - \sin d \cos c \\ &= 2 \sin c \cos d \\ &= 2 \sin\left(\frac{X+Y}{2}\right)\cos\left(\frac{X-Y}{2}\right)\end{align}$$
Making $X = ax$ and $Y = (a+2)x$ gives: $$\sin(ax) + \sin((a+2)x) = 2\sin((a+1)x))\cos(-x) = 2\sin((a+1)x)~\cos(x)$$
since $\cos$ is an even function.
