1. Determine all pairs $(x, y)$ of real numbers for which

$$
\log _{2}\left[2 \cos ^{2}(x y)+\frac{1}{2 \cos ^{2}(x y)}\right]-1=-\left(y-\frac{1}{2}\right)^{2}
$$
1. Since $2 \cos ^{2}(x y)$ and $\frac{1}{2 \cos ^{2}(x y)}$ are positive numbers, we can apply the AM-GM inequality:

$$
2 \cos ^{2}(x y)+\frac{1}{2 \cos ^{2}(x y)} \geq 2 \sqrt{2 \cos ^{2}(x y) \cdot \frac{1}{2 \cos ^{2}(x y)}}=2 . \quad 5 \text { points }
$$

From this, it follows that,

$$
\log _{2}\left[2 \cos ^{2}(x y)+\frac{1}{2 \cos ^{2}(x y)}\right]-1 \geq \log _{2} 2-1=0
$$

On the other hand, $-\left(y-\frac{1}{2}\right)^{2} \leq 0$. Therefore, for the equality to hold, it must be that $y=\frac{1}{2}$ and

$$
2 \cos ^{2}(x y)=\frac{1}{2 \cos ^{2}(x y)}
$$

This implies,

$$
2 \cos ^{2} \frac{x}{2}=\frac{1}{2 \cos ^{2} \frac{x}{2}}
$$

From this, we get

$$
\cos ^{4} \frac{x}{2}=\frac{1}{4} \Rightarrow \cos ^{2} \frac{x}{2}=\frac{1}{2} \Rightarrow \frac{1+\cos x}{2}=\frac{1}{2}
$$

i.e., $\cos x=0$ and $x=\frac{\pi}{2}+k \pi, \quad k \in \mathbf{Z}$.

Therefore, $(x, y) \in\left\{\left(\frac{(2 k+1) \pi}{2}, \frac{1}{2}\right): k \in \mathbf{Z}\right\}$.

10 points