Task B-3.4. Determine the solutions of the inequality:

$$
2011 \cos \left(2 x^{2}-y\right) \geq x^{2}+2011
$$
Solution. From $x^{2} \geq 0$ it follows that $x^{2}+2011 \geq 2011$, and from $\cos \left(2 x^{2}-y\right) \leq 1$ it follows that

$$
2011 \cos \left(2 x^{2}-y\right) \leq 2011
$$

We conclude that the only possible solution is when $2011 \cos \left(2 x^{2}-y\right)=x^{2}+2011=2011$, from which we get $x^{2}=0$ so $x=0$.

Furthermore, $1=\cos (-y)=\cos y$ so $y=2 k \pi, k \in \mathbb{Z}$.

Thus, the solution is $\{(0,2 k \pi): k \in \mathbb{Z}\}$.