## Task A-3.3.

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

$$
\sin \left(x^{2}-5 y\right)-1 \geqslant \frac{x^{4}}{2016}
$$
## Solution.

Since the value of sine is at most 1, it holds that $\sin \left(x^{2}-5 y\right)-1 \leqslant 0$.

Since the squares of real numbers are non-negative, it holds that $\frac{x^{4}}{2016} \geqslant 0$.

1 point

It follows that both sides must be exactly equal to 0.

Therefore, $x=0$. 1 point

Then, from $\sin (-5 y)=1$ it follows that $-5 y=\frac{\pi}{2}+2 k \pi$, and then $y=-\frac{\pi}{10}-\frac{2 k \pi}{5}$, for $k \in \mathbb{Z}$. 1 point The set of solutions is therefore $\left\{\left.\left(0,-\frac{\pi}{10}-\frac{2 k \pi}{5}\right) \right\rvert\, k \in \mathbb{Z}\right\}$.