5. Solve the system of equations:

$$
\left\{\begin{array}{l}
x y=1 \\
x+y+\cos ^{2} z=2
\end{array}\right.
$$

Answer: $x=y=1 ; z=\frac{\pi}{2}+\pi n$, where $n \in Z$.
Solution. From the first equation, $x, y$ are both positive or both negative. But from the second equation $x+y \geq 1$, so they are both positive. Then, applying the inequality of means and using the first equation of the system, we get: $x+y \geq 2 \sqrt{x y}=2$. Since $\forall z \in R \cos ^{2} z \geq 0$, it follows from the second equation that $x+y=2 u \cos z=0$. Hence $x=y=1 ; z=\frac{\pi}{2}+\pi n$, where $n \in Z$

Remark. We could have avoided using the inequality of means and instead replaced 2 in the second equation with $2xy$ (which is possible since $xy=1$), moved it to the other side, and completed the square.