I was doing 1. (C) and ran into the problem of one of my critical numbers being a negative root. What do I do with that? I thought you mentioned something about it in class, but I couldn't remember.

I am assuming you meant "(sqaure) root of a negative number" as opposed to "negative root." In this course, we stick to the context of the real numbers. You can do calculus with complex numbers, but we'll avoid them. So, if you encounter an even root of a negative number, we would say that there is no (real) solution.

Also on 1. (D) How do I solve for the critical points of: $K'(x)=-2sin(x)+cos(2x)$

First, some trivial comments:

1. You mean lowercase $k$ as opposed to $K$. Of course, you should use a capital letter if that's what the original function was called.

2. To typeset sine and cosine using $\LaTeX$, you should use `\sin` and `\cos`. Notice the difference between $-2sin(x)+cos(2x)$ and $-2\sin(x)+\cos(2x)$. The last expression was typeset using

OK, now to solve for the critical numbers, the easiest thing to do is make use of the following identity:

(1)
\begin{align} \cos(2x)=1-2\sin^2(x) \end{align}

I'm not expecting you to have this identity (or its two other brothers), but you should know that it exists and where to find it. The reason why this particular identity is useful is that then the whole equation can be written in terms of $\sin(x)$ and in this case, you can factor because it is quadratic in form.