How to solve these two trigonometric equations? How to solve these two trigonometric equations :

$$\sin y \sin(2x+y)=0$$
  $$\sin x \sin(x+2y)=0.$$

I know one set of solution will be $(0,0)$. What will be the other set ?
 A: A possibility is $\sin y=0$, which gives $y=k\pi$ (integer $k$); substituting in the second one gives
$$
\sin x\sin(x+2k\pi)=0
$$
so this implies $\sin x=0$. Similarly, $\sin x=0$ implies $\sin y=0$. Thus we can reduce to the case
$$\begin{cases}
\sin(2x+y)=0\\
\sin(x+2y)=0
\end{cases}
$$
From this you have
$$
\begin{cases}
2x+y=a\pi\\
x+2y=b\pi
\end{cases}
$$
(where $a$ and $b$ are any integers).
This is a linear system, that can be solved as
$$
x=\frac{2a-b}{3}\pi,\quad
y=\frac{2b-a}{3}\pi
\qquad(a,b\text{ any integers}).
$$
This formula encompasses also the case $\sin x=\sin y=0$, because given arbitrary integers $h$ and $k$ one can express
$$
\begin{cases}
h=\frac{2a-b}{3}\\[1ex]
k=\frac{2b-a}{3}
\end{cases}
$$
for suitable integers $a$ and $b$, namely $a=2h+k$ and $b=h+2k$.
Therefore the general solution can be written as
$$
\boxed{\displaystyle x=\frac{2a-b}{3}\pi,\quad y=\frac{2b-a}{3}\pi,
\qquad a,b\text{ arbitrary integers}}.
$$

Noticing that $2a-b\equiv 2b-a\pmod{3}$, we can express the solutions in a different form, as three families:
\begin{align}
&1. & x &= h\pi,                & y &= k\pi\\[1ex]
&2. & x &= \frac{\pi}{3}+h\pi,  & y &= \frac{\pi}{3}+k\pi\\[1ex]
&3. & x &= \frac{2\pi}{3}+h\pi, & y &= \frac{2\pi}{3}+k\pi
\end{align}
again for arbitrary integers $h$ and $k$.
A: From the first equation
Case $1:$ If $\sin y=0,y=n\pi$  where $n$ is any integer
Assuming the second equation to be $\sin x\sin(x+2y)=0$
$\sin x\sin(x+2y)=\sin x\sin(x+2n\pi)=\sin^2x$
$\implies \sin^2x=0\implies \sin x=0\implies x=m\pi$  where $m$ is any integer
So, one set of solution is $(x,y)=(m\pi,n\pi)\ \ \  \ (1)$
Case $2:$ If $\sin(2x+y)=0,2x+y=r\pi$  where $r$ is any integer
From the second equation, $\sin x\sin(x+2y)=\sin x\sin(x+2r\pi-4x)=-\sin x\sin3x$
$\implies \sin x\sin3x=0$
If $\sin x=0, x=p\pi,$ where $p$ is any integer
$\implies y=r\pi-2x=r\pi-2p\pi=(r-2p)\pi\ \ \ \ (2)$  which is clearly a subset of $(1)$
If $\sin3x=0,3x=q\pi$  where $q$ is any integer
$\implies  x=\frac{q\pi}3$
and subsequently $y=r\pi-2x=r\pi-2\frac{q\pi}3=\frac{(3r-2q)\pi}3$
So, another set of solution is $(x,y)=\left(\frac{q\pi}3,\frac{(3r-2q)\pi}3\right)\ \ \  \ (3)$ 
If we compare $(1),(3)$
$m\pi=\frac{q\pi}3\implies q=3m$  and $n\pi=\frac{(3r-2q)\pi}3=\frac{(3r-2\cdot3m)\pi}3=(r-2m)\pi\implies r=n+2m$
Do you see the relationship between $(1),(3)?$ Essentially, the solution can be expressed as one set, right? How?
