How would you determine $\sin(x) = -\cos(x)$ I can pretty much look at the equation and say the answer is $\frac{3\pi}{4} + 2\pi k$ and $\frac{7\pi}{4} + 2\pi k$ (where $k \in \mathbb N$) but how can I show work for this?
 A: Divide both sides by $\cos x$ to get $\tan x = -1$.
A: This may be overkill, but it could be useful in different situations, for instance something like $\sin x=-\cos3x$.
Recall that $-\cos x=\sin(x-\pi/2)$ so your equation becomes
$$
\sin x=\sin\left(x-\frac{\pi}{2}\right)
$$
This amounts to saying that either
$$
x-\frac{\pi}{2}=x+2k\pi
$$
(which has no solutions) or
$$
x-\frac{\pi}{2}=\pi-x+2k\pi
$$
that becomes
$$
2x=\pi+\frac{\pi}{2}+2k\pi
$$
and, finally,
$$
x=\frac{3}{4}\pi+k\pi
$$
A: or we divide by $\sin(x)$ and we get $$\cot(x)=-1$$
A: Good question, so +1. Joel Cohen's idea is good, it is always a good idea to reduce the amount of work one needs to do. 
Another way of answering it (that shows understanding of the periodic properties of trig functions) is by noting $$\sin(x + k\pi) = (-1)^k \sin(x)$$ and $$ \cos(x + k\pi) = (-1)^k \cos(x).$$
The above facts mean that we only need to seek solutions (to your equation) $a$, where $ 0 \leq a < \pi$. The only solution in this range is $\frac{ 3\pi}{4}$.
So some solutions are: $ \frac{3\pi}{4} + k \pi $, for any $k$, an integer. They are in fact all the solutions, as the trig functions are $2\pi$ periodic.
