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If I have a system of period functions of $x$ and $y$, in this case trigonometric,

$$\begin{cases} \sin{(2x + y)} = 0 \\ \sin{(2y + x)} = 0 \end{cases}$$

is it okay for me to to use the same variable $k$ to describe the period at which those two conditions are met, or should I use two separate values?

$$\begin{cases} 2x + y = k \pi \\ 2y + x = k \pi \end{cases}, k \in \mathbb{Z}$$

I would think that using the same variable is fine, since the important aspect is to describe the period at which the original conditions hold true, not specifically how many times you have to travel $\pi$ distance around the circle in each case.

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not for me! :-) You seem to imply that $2x+y=2y+x$ – Raymond Manzoni Jan 22 '12 at 15:23
up vote 3 down vote accepted

You need to use two different variables, because you're not done yet -- the natural next step would be to solve the resulting system of two linear equations, which would give you explicit expressions for $x$ and $y$ in terms of the $k$'s. But each of those expressions would contain both $k$'s, so you'd need some way to tell them apart during your computations.

Otherwise, you would be assuming (with no good reason) that $k$ is the same for both original equations -- in other words that $2x+y=2y+x$ -- something that is not implied by your original equations. As a result you would miss most of their solutions.

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No, I don't think it would be fine to use the same parameter.

You are solving a trigonometric equation. So, sines vanish at all integral multiples of $\pi$. By claiming that both $2x+y$ and $2y+x$ equal $k\pi$, you're neglecting many solutions to the system.

How is that?

That is because, say, $2y+x= 57\pi$ and $2x+y=17 \pi$, this satisfies the system, but, your parametric solution doesn't count it!

So, what do I do?

It's just simple, use different parameters for $2x+y$ and $2y+x$.

So, your solution should look like,

$$2x+y=m\pi\tag{1}$$ $$2y+x=n \pi\tag{2}$$

FWIW, I will add a trick that comes to mind:

Add (1) and (2), $$x+y=\dfrac{(m+n)\cdot \pi}{3}\tag{3}$$ and, subtract (1) and (2), you'll get $$x-y=(m-n)\cdot \pi\tag{4}$$

Now for the finishing touch, add (3) and (4) for $x$ and subtract them for $y$.

See, if you get, $$\boxed{x=\dfrac{\pi(2m-n)}{3}}$$ and $$\boxed{y=\dfrac{\pi(2n-m)} {3}}$$

I thought I'll add this last piece, using same variables means assuming that $2x+y=2y+x$ for no reason you and I can think of! So, it's essentially not a good idea to do so. As I have previously written, it leads to loss of a good number of solutions!

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