How to solve these two trigonometric equations?

Question

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 ?

Answer 1

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 $$ \boxed{\displaystyle x=\frac{2a-b}{3}\pi,\quad y=\frac{2b-a}{3}\pi, \qquad a,b\text{ arbitrary integers}}. $$

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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 $k$.

Answer 2

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?