Solution:
For $x$ to be such a number is equivalent to $x$ being a $k^{\text{th}}$ root of unity for some $k$ up to $2012$. For each $k$, there are $\varphi(k)$ primitive $k^{\text{th}}$ roots of unity, so the total number of roots is $\sum_{k=1}^{2012} \varphi(k)$.

We will give a good approximation of this number using well known facts about the Möbius function, defined by $\mu(n)=\left\{\begin{array}{ll}0 & \text{ if } n \text{ is not squarefree } \\ (-1)^{r} & \text{ if } n \text{ has } r \text{ distinct prime factors }\end{array}\right.$. It turns out that if $f(n)=\sum_{d \mid n} g(d)$, then $g(n)=\sum_{d \mid n} \mu(d) f\left(\frac{n}{d}\right)$. Using this fact, since $n=\sum_{d \mid n} \varphi(d)$, we have that $\varphi(n)=\sum_{d \mid n} \mu(d) \frac{n}{d}$. Now we have reduced the problem to estimating $\sum_{k=1}^{2012} \sum_{d \mid k} \mu(d) \frac{k}{d}$. Let $a=\frac{k}{d}$, so we obtain $\sum_{k=1}^{2012} \sum_{d \mid k} a \mu(d)$.

We can interchange the order of summation by writing
$$\begin{aligned} \sum_{d=1}^{2012} \sum_{a=1}^{\left\lfloor\frac{2012}{d}\right\rfloor} a \mu(d) & \approx \sum_{d=1}^{2012} \mu(d) \frac{1}{2}\left(\left\lfloor\frac{2012}{d}\right\rfloor\right)^{2} \\ & \approx \sum_{d=1}^{2012} \mu(d) \frac{2012^{2}}{2 d^{2}} \\ & =\frac{2012^{2}}{2} \sum_{d=1}^{2012} \frac{\mu(d)}{d^{2}} \\ & \approx \frac{2012^{2}}{2} \sum_{d=1}^{\infty} \frac{\mu(d)}{d^{2}} \end{aligned}$$
The Möbius function also satisfies the property that $\sum_{d \mid n} \mu(d)=\left\{\begin{array}{ll}1 & \text{ if } n=1 \\ 0 & \text{ otherwise }\end{array}\right.$, which can be seen as a special case of the theorem above (letting $f(n)=1, g(n)=\left\{\begin{array}{ll}1 & \text{ if } n=1 \\ 0 & \text{ otherwise }\end{array}\right.$ ). We can then see that $\left(\sum_{d=1}^{\infty} \frac{\mu(d)}{d^{2}}\right)\left(\sum_{c=1}^{\infty} \frac{1}{c^{2}}\right)=\frac{1}{1^{2}}=1$, so $\sum_{d=1}^{\infty} \frac{\mu(d)}{d^{2}}=\frac{6}{\pi^{2}}$. Therefore, we have $\sum_{k=1}^{2012} \varphi(k) \approx \frac{3}{\pi^{2}} \cdot 2012^{2}=$ 1230488.266... 2012 is large enough that all of our approximations are pretty accurate and we should be comfortable perturbing this estimate by a small factor to give bounding values.