Solving $\sin{3x} = \sin{x}$: question about the $k2\pi$ part Why is $$\sin{3x} = \sin{x}$$ 
equivalent to 
$$3x = x + k2\pi$$
$$3x = \pi - x + k2\pi$$
and not 
$$3x + k2\pi = x + k2\pi$$
$$3x + k2\pi = \pi - x + k2\pi$$
 A: 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 $m$.
A: When you have $\sin\alpha=\sin\beta$, there are two cases:


*

*$\alpha$ and $\beta$ differ by an integral multiple of $2\pi$, or

*$\alpha$ 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
$$
x=\frac{\pi}{4}+k\frac{\pi}{2} \qquad (k\in\mathbb{Z})
$$
A: 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*}
