Solving $\sin{3x} = \sin{x}$: question about the $k2\pi$ part

Question

Why is $$\sin{3x} = \sin{x}$$

equivalent to

$$3x = x + k2\pi$$ $$3x = \pi - x + k2\pi$$

and not

$$3x + k2\pi = x + k2\pi$$ $$3x + k2\pi = \pi - x + k2\pi$$

Answer 1

Great question!

What you really have is

$$ 3x+k2\pi=x+p2\pi, $$

where $k$ and $p$ are arbitrary integers. (You should use different variables to make it clear that they needn't be the same integer).

Now, moving one term to the other side yields:

$$ 3x=x+(p-k)2\pi. $$

Since $p$ and $k$ were arbitrary, $p-k$ will just be an arbitrary integer. You can then replace $p-k$ by just one letter, m, say, and conclude

$$ 3x=x+m2\pi, $$

for any integer $m$.

Answer 2

When you have $\sin\alpha=\sin\beta$, there are two cases:

1. $\alpha$ and $\beta$ differ by an integral multiple of $2\pi$, or
2. $\alpha$ and $\pi-\beta$ differ by an integral multiple of $2\pi$.

The first case can be written $\alpha-\beta=2k\pi$ or $\alpha=\beta+2k\pi$ (where $k$ is an integer).

You might write $\alpha+2h\pi=\beta+2k\pi$, with $h$ and $k$ integers, but this becomes $\alpha=\beta+2(k-h)\pi$. Note that it would be wrong to write $\alpha+2k\pi=\beta+2k\pi$, because there's no relation between the two integers above. However, since $h$ and $k$ are arbitrary, also $k-h$ is an arbitrary integer as well, so there's no real need to do this way.

Similarly, the second case can be written $\alpha-(\pi-\beta)=2k\pi$ for an integer $k$, or $\alpha=\pi-\beta+2k\pi$.

Thus your equation can be solved by $$ 3x=x+2k\pi \qquad (k\in\mathbb{Z}) $$ or $$ 3x=\pi-x+2k\pi \qquad (k\in\mathbb{Z}) $$ Easy algebraic manipulations give the compact form for the solutions: $$ x=k\pi \qquad (k\in\mathbb{Z}) $$ or $$ x=\frac{\pi}{4}+k\frac{\pi}{2} \qquad (k\in\mathbb{Z}) $$

Answer 3

It's much simpler using the language of congruences: \begin{align*} \sin\alpha=\sin\beta\iff\begin{cases}\alpha\equiv\beta\\\text{or}\\\alpha\equiv\pi-\beta\end{cases}\mod2\pi, \end{align*} and similarly \begin{alignat*}{2} &\cos\alpha=\cos\beta&&\iff\alpha\equiv\pm\beta\mod2\pi,\\ &\tan\alpha=\tan\beta&&\iff\alpha\equiv\beta\mod\pi. \end{alignat*}