Task B-3.2. Solve the inequality

$$
\cos (2010 x+y) \geq y^{4}-2 y^{2}+2
$$
## Solution.

The right side of the given inequality is the function $f(y)=y^{4}-2 y^{2}+2$. Let's write $f$ in the following form:

$$
f(y)=\left(y^{2}-1\right)^{2}+1
$$

Then $f(y) \geq 1, \forall y \in \mathbb{R}$.

On the other hand, $\cos (2010 x+y) \leq 1, \forall x, y \in \mathbb{R}$.

We conclude that the given inequality can have a solution only if $\left(y^{2}-1\right)^{2}+1=1$ and if $\cos (2010 x+y)=1$.

From the first equality, we get that $y= \pm 1$.

From $\cos (2010 x+y)=1$, we get $2010 x+y=2 k \pi$.

Then for $y= \pm 1, x=\mp \frac{1}{2010}+\frac{k \pi}{1005}, k \in \mathbb{Z}$.