1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& y+3 x=4 x^{3}, \\
& x+3 y=4 y^{3} .
\end{aligned}
$$
Let $(x, y)$ be a solution of the given system. If $|x|>1$, the first equation yields $|y|=|x|\left|4 x^{2}-3\right|>|x|$, since $4 x^{2}-3>1$. We will then have $|y|>|x|>1$ and from the second equation, due to the derived inequality $|y|>1$, we can similarly deduce $|x|>|y|$, which leads to a contradiction. Therefore, $-1 \leq x \leq 1$ and in the interval $[0, \pi]$ there exists $t$ such that $x=\cos t$. Substituting into the first equation, we get $y=4 \cos ^{3} t-3 \cos t=\cos 3 t$. From the second equation, we similarly get $x=4 \cos ^{3} 3 t-3 \cos 3 t=\cos 9 t$. Therefore, $\cos t=\cos 9 t$, and after rearrangement,

\[
\begin{aligned}
\cos t-\cos 9 t & =0 \\
2 \sin \frac{9 t+t}{2} \sin \frac{9 t-t}{2} & =0 \\
\sin 5 t \sin 4 t & =0
\end{aligned}
\]

The last equality is satisfied precisely when $5 t$ or $4 t$ is a multiple of $\pi$. Together with the condition $0 \leq t \leq \pi$, this gives solutions of the form $(\cos t, \cos 3 t)$ for

\[
t \in\left\{0, \frac{1}{5} \pi, \frac{1}{4} \pi, \frac{2}{5} \pi, \frac{1}{2} \pi, \frac{3}{5} \pi, \frac{3}{4} \pi, \frac{4}{5} \pi, \pi\right\}.
\]

In this case, verification is also not necessary.