Solve the following system of equations:

$$
\begin{gathered}
x^{2}+y^{2}=1 \\
4 x^{3}-3 x=\sqrt{\frac{y+1}{2}}
\end{gathered}
$$
The equation $x^{2}+y^{2}=1$ represents a unit circle centered at the origin; therefore, there exists a number $\varphi \in[0,2 \pi)$ such that

$$
x=\cos \varphi \quad \text { and } \quad y=\sin \varphi
$$

Substituting these into the second equation:

$$
4 \cos ^{3} \varphi-3 \cos \varphi=\sqrt{\frac{\sin \varphi+1}{2}}
$$

The left side of the equation is exactly $\cos 3 \varphi$; thus, the equation can be rewritten as:

$$
\cos 3 \varphi=\sqrt{\frac{\sin \varphi+1}{2}}
$$

By ensuring that $\cos 3 \varphi \geq 0$, we square both sides and rearrange the equation:

$$
\begin{gathered}
2 \cos ^{2} 3 \varphi-1=\sin \varphi \\
\cos 6 \varphi=\cos \left(\frac{\pi}{2}-\varphi\right)
\end{gathered}
$$

This is possible if $6 \varphi=(\pi / 2-\varphi)+2 k \pi$, or $6 \varphi=-(\pi / 2-\varphi)+2 k \pi$, that is,

$$
\varphi=\frac{\pi}{14}+\frac{2 k \pi}{7} \quad \text { or } \quad \varphi=-\frac{\pi}{10}+\frac{2 k \pi}{5}
$$

From this, considering that $0 \leq \varphi<2 \pi$ and $\cos 3 \varphi \geq 0$, we obtain the following values:

$$
\begin{array}{lll}
\varphi_{1}=\frac{\pi}{14} ; & \varphi_{2}=\frac{9 \pi}{14} ; & \varphi_{3}=\frac{17 \pi}{14}, \quad \text { and } \\
\varphi_{4}=\frac{7 \pi}{10} ; & \varphi_{5}=\frac{3 \pi}{2} ; & \varphi_{6}=\frac{19 \pi}{10}
\end{array}
$$

The corresponding solutions are:

$$
\begin{array}{llll}
x_{1}=\cos \frac{\pi}{14}, & y_{1}=\sin \frac{\pi}{14} ; & x_{4}=\cos \frac{7 \pi}{10}, & y_{4}=\sin \frac{7 \pi}{10} \\
x_{2}=\cos \frac{9 \pi}{14}, & y_{2}=\sin \frac{9 \pi}{14} ; & x_{5}=0, & y_{5}=-1 \\
x_{3}=\cos \frac{17 \pi}{14}, & y_{3}=\sin \frac{17 \pi}{14} ; & x_{6}=\cos \frac{19 \pi}{10}, & y_{6}=\sin \frac{19 \pi}{10}
\end{array}
$$