351*. Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
4 x y\left(2 y^{2}-1\right)=1
\end{array}\right.
$$
$\triangle$ Given the first equation of the system, we introduce trigonometric substitutions

$$
x=\cos \alpha, \quad y=\sin \alpha
$$

where $\alpha \in[0 ; 2 \pi]$. We transform the second equation:

$4 \cos \alpha \sin \alpha \cos 2 \alpha=1, \quad \sin 4 \alpha=1$.

Then

$$
4 \alpha=\frac{\pi}{2}+2 \pi k, \quad \alpha=\frac{\pi}{8}+\frac{\pi k}{2}(k \in Z)
$$

Since $\alpha \in[0 ; 2 \pi]$, we should limit ourselves to the values $k=0,1,2,3$. We obtain:

$$
\alpha_{1}=\frac{\pi}{8}, \quad \alpha_{2}=\frac{5 \pi}{8}, \quad \alpha_{3}=\frac{9 \pi}{8}, \quad \alpha_{4}=\frac{13 \pi}{8}
$$

Answer: $(\cos \pi / 8 ; \sin \pi / 8),(\cos 5 \pi / 8 ; \sin 5 \pi / 8),(\cos 9 \pi / 8 ; \sin 9 \pi / 8),(\cos 13 \pi / 8$; $\sin 13 \pi / 8)$.