<s>[INST] Trouble Calculating for $30 \cos(30x)+14=−16$ I am unable to see how,
$$30\cos(30x)+14=−16$$
is equal to
$$\frac{\pi}{30} + n \frac{\pi}{15}$$
I solved up to this
$$\cos(30x) = -1$$
$$\pi= 30x$$
$$\frac{\pi}{30} = x$$
But I am unsure where the $\frac{\pi}{15}$ came from. Can someone help me understand this?
 [/INST] The thing is $\cos (x + 2n\pi ) = \cos x$ so if $\cos 30x = -1$ then $\cos 30(x+ \frac {2n\pi}{30})$ will also be equal to $-1$.
So when you got $\cos (30x ) = -1$ then 
$30x = \pi$ is ONE of the possible values for $30x$ (because $\cos \pi = -1$).
$30x = 3\pi$ is another one (because $\cos 3\pi = -1)$ and $30x = -pi$ and $30x=5\pi$ and $30x = 2147\pi$ are some more (because $\cos -\pi = \cos 5\pi = \cos 2147 \pi = -1$).
Indeed for any integer, $n$, you have an infinite number of possibilities of $30x = (2n+1)\pi$ becase $\cos (2n+1)\pi = -1$.
So $x =\frac {\pi}{30}$ is one answer and $\frac {\pi}{10}$ or $-\frac {\pi}{30}$ and $\frac {\pi}{6}$ and $\frac {2147\pi}{30}$ are all also solutions because for all thus values of $x$ you'd have $30x = \{\pi,3\pi, -pi,5\pi, 2147\pi\}$ and $\cos 30x = -1$ for all of those.
In fact, there are an infinite number of solutions.  For any $\frac{\pi}{30} + \frac n{15}\pi$ you will have $\cos 30x = \cos (30(\frac {\pi}{30} + \frac n{15}\pi))=\cos (\pi + 2n \pi) = \cos \pi = -1$.
So those are an infinite number of solutions where $\cos 30x = -1$
</s>