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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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up vote 4 down vote accepted

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.

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My best guess is that $k$ depends on the function with the highest amount of (half)cycles on the given interval? Is that correct? – Martndemus Jul 16 '12 at 16:31

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