Why doesn't $\arccos x = -\tfrac12\sqrt{3}$ have any solutions? I have this exercise with an unclear answer. The question is this:
$$\arccos x = -\frac{\sqrt3}{2}\,.$$
The answer is this:
$$\begin{gather*}
\varphi(x)= \arccos x\\
V_\varphi = [0,\pi]\\
-\frac{\sqrt3}{2}\notin V_\varphi\,.
\end{gather*}
$$
can someone this a bit more to me? How does this prove that there are no solutions?

As a couple of others have already pointed out, $-\sqrt{3}/2$ is simply not in the range of the arccosine.  Here's an explanation as to why that's true.
Here's the graph of the cosine function over the interval $[-\pi,2\pi]$:

The issue is that this function is not one-to-one.  As a result, we must restrict it to an appropriate domain where it is one-to-one in order to talk a about a restricted inverse function.  Conventionally, the interval $[0,\pi]$ is chosen, which yields something like the following:

Now, the inverse of this restricted version of the cosine is what we know as the arccosine and its graph looks like so:

Of course, the domain and range have flipped - thus, $-\sqrt{3}/2$ (or any other negative number) is not in the range!
