## k in trig equality [closed]

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 $[-2,2]$?

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This question is not on-topic on this site whose scope is quite narrow, see the FAQs for details. It might be on-topic on math.stackexchange.com a similar site with a broader scope. – quid Jul 16 at 15:21