Answer: $x = \frac{3}{2} + 2n, n \in \mathbb{Z}$.

Since $a - [a] = \{a\}$, where $\{a\}$ is the fractional part of $a$, we can rewrite our equation in the following way:
$$\cos \pi x = \left[ \frac{\{x\}}{2} - \frac{1}{2} \right].$$
Obviously, $0 \leq \frac{\{x\}}{2} < 1$ for every real $x$. Consider two cases:

1) Let $0 \leq \frac{\{x\}}{2} < \frac{1}{2}$. Then $-\frac{1}{2} \leq \frac{\{x\}}{2} - \frac{1}{2} < 0$, and thus $[\frac{\{x\}}{2} - \frac{1}{2}] = -1$. So in this case we get the equation $\cos \pi x = -1$. The solutions of this equation are $x = 1 + 2k, k \in \mathbb{Z}$. But for such $x$ we have that $\{\frac{x}{2}\} = \{\frac{1}{2} + k\} = \frac{1}{2}$, which contradicts our assumption. So, we obtain that there are no solutions in this case.

2) Let $\frac{1}{2} \leq \frac{\{x\}}{2} < 1$. Then $0 \leq \frac{\{x\}}{2} - \frac{1}{2} < \frac{1}{2}$, which implies that $[\frac{\{x\}}{2} - \frac{1}{2}] = 0$. So, in this case our equation reduces to the equation $\cos \pi x = 0$. The solutions for this equation are $x = \frac{1}{2} + k$, $k \in \mathbb{Z}$. For such $x$ we have:
$$\begin{cases} \frac{x}{2} \\ \frac{1}{2} \end{cases} = \begin{cases} \frac{1}{4} + k \\ \frac{1}{2} \end{cases} = \begin{cases} \frac{1}{4}, & k = 2n, \\ \frac{3}{4}, & k = 2n+1. \end{cases}$$
So, for $\frac{1}{2} \leq \frac{\{x\}}{2} < 1$ we should take $k = 1 + 2n$, $n \in \mathbb{Z}$. Therefore, $x = \frac{3}{2} + 2n$, $n \in \mathbb{Z}$.