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 $[-2,2]$?
