$k$ in trigonometric equality $\sin(a) =\sin(b)$ On a test there is the question: "Solve for $x$ on the interval $[-\pi,\pi]$ where $\sin(2x) = \cos(3x)$
I know that: 
$\cos(x) = \sin(\frac12\pi - x)$ 
So you can rewrite the equation to: 
$\sin(2x) = \sin(\frac12\pi - 3x)$
But then in the solution, the next step is this: 
$2x = \frac12 \pi - 3x + 2\pi k$ or $2x = \pi - (\frac12\pi - 3x) + 2\pi k$
What is the $2\pi k$ for? 
Later they simplify it to:
$x = \frac{1}{10}\pi + \frac25\pi k$ or $x = -\frac12\pi + 2\pi k $
and then it goes like this:
$x = \frac{1}{10}\pi - 2 * \frac25\pi = -\frac7{10}\pi$
$x = -\frac12\pi $
$x = \frac1{10}\pi - 1 * \frac25\pi = -\frac3{10}\pi$ 
$x = \frac1{10}\pi $
$x = \frac{1}{10}\pi + 1 * \frac25\pi = \frac12\pi$
$x = \frac{1}{10}\pi + 2 * \frac25\pi = \frac9{10}\pi$
How does that part work? I can't find any theory on it. Why is $k$ substituted by the range $[-2,2]$?
 A: Both $\sin$ and $\cos$ are periodic with a period of $2\pi$. This means that $$\ldots=\sin(x-2\pi-2\pi)=\sin(x-2\pi)=\sin(x)=\sin(x+2\pi)=\sin(x+2\pi+2\pi)=\ldots$$
Namely, for every integer (positive, negative or zero) we have $\sin(x)=\sin(x+2\pi k)$, and similarly for $\cos$.
When we are solving an equality such as $\cos(x)=1$ then the results are $x=0,2\pi,4\pi,6\pi,\ldots$ but if the question specified $x$ is in a particular interval, then one can calculate the exact value of $k$ (or several possible values).
For example, $\sin(x)=0$ for $x\in[-\frac\pi2,\frac\pi2]$ then we know that $x=0$, since the general solution is $x=2\pi k$ but for $k\neq 0$ we have that $2\pi k\notin[-\frac\pi2,\frac\pi2]$.
In your question, once you know that $x\in[-\pi,\pi]$ then the value of $k$ is determined:
If $2x=\frac12\pi-3x+2\pi k$, we can instead just write $x=\frac1{10}\pi+\frac4{10}\pi k$. The only values of $k$ for which $x\in[-\pi,\pi]$ are $k=\pm1,\pm2,0$. 
The other case, I leave you to discover the values of $k$ alone.
