11.4. Solve the system of equations

$$
\left\{\begin{array}{l}
x \sqrt{1-y^{2}}=\frac{1}{4}(\sqrt{3}+1) \\
y \sqrt{1-x^{2}}=\frac{1}{4}(\sqrt{3}-1)
\end{array}\right.
$$
# Solution:

Method 1. From the domain of the system, it follows that $-1 \leqslant x, y \leqslant 1$. Moreover, both these numbers are positive, since the right-hand sides of the equations are positive. Therefore, we can assume that $x=\sin \alpha, y=\sin \beta$ for some $\alpha$ and $\beta$ from the interval $\left(0 ; \frac{\pi}{2}\right]$. The system then takes the form

$$
\left\{\begin{aligned}
\sin \alpha \cdot \cos \beta & =\frac{1}{4}(\sqrt{3}+1) \\
\sin \beta \cdot \cos \alpha & =\frac{1}{4}(\sqrt{3}-1)
\end{aligned}\right.
$$

By adding the left and right parts of the equations, we get $\sin (\alpha+\beta)=\frac{\sqrt{3}}{2}$, and by subtracting, we find that $\sin (\alpha-\beta)=\frac{1}{2}$. Since the number $\alpha-\beta$ lies on the interval $\left[-\frac{\pi}{2} ; \frac{\pi}{2}\right]$,
it equals $\frac{\pi}{6}$. The number $\alpha+\beta$ lies on the interval $[0 ; \pi]$, and therefore equals either $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$. Then $\alpha=\frac{(\alpha+\beta)+(\alpha-\beta)}{2}$ is either $\frac{5 \pi}{12}$ or $\frac{\pi}{4}$. The corresponding values of $\beta$ are $\frac{\pi}{4}$ and $\frac{\pi}{12}$, respectively. We have two pairs $\left(\sin \frac{5 \pi}{12} ; \sin \frac{\pi}{4}\right)$ and $\left(\sin \frac{\pi}{4} ; \sin \frac{\pi}{12}\right)$.

Method 2. Squaring both equations, which is an equivalent transformation for $0 \leqslant x, y \leqslant 1$, and substituting $x^{2}=a, y^{2}=b$, we obtain the system

$$
\left\{\begin{array}{l}
a(1-b)=\frac{2+\sqrt{3}}{8} \\
b(1-a)=\frac{2-\sqrt{3}}{8}
\end{array}\right.
$$

Subtracting the second equation from the first, we find $a-b=\frac{\sqrt{3}}{4}$. Expressing $a$ from this and substituting this value into the first equation of the system, we get

$$
\left(b+\frac{\sqrt{3}}{4}\right) \cdot(1-b)=\frac{2+\sqrt{3}}{8}
$$

Solving this quadratic equation, we find $b_{1}=\frac{1}{2}, b_{2}=\frac{2-\sqrt{3}}{4}$. Since we have $a=b+\frac{\sqrt{3}}{4}$, we get $a_{1}=\frac{\sqrt{3}+2}{4}, a_{2}=\frac{1}{2}$. Considering the non-negativity of the numbers $x$ and $y$, we find two solutions to the system: $\left(\frac{\sqrt{2}}{2} ; \frac{\sqrt{2-\sqrt{3}}}{2}\right)$ and $\left(\frac{\sqrt{2+\sqrt{3}}}{2} ; \frac{\sqrt{2}}{2}\right)$

Answer: $\left\{\left(\sin \frac{5 \pi}{12} ; \sin \frac{\pi}{4}\right),\left(\sin \frac{\pi}{4} ; \sin \frac{\pi}{12}\right)\right\}$. Another form of writing: $\left\{\left(\frac{\sqrt{2+\sqrt{3}}}{2} ; \frac{\sqrt{2}}{2}\right),\left(\frac{\sqrt{2}}{2} ; \frac{\sqrt{2-\sqrt{3}}}{2}\right)\right\}$.

| in progress | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| A correct answer is obtained with errors in arithmetic or when working with square roots | 5 points |
| Only one pair $(x, y)$ solving the system is correctly and justifiedly found | 3 points |
| A trigonometric substitution is made or a correct quadratic equation is obtained relative to one of the variables | 2 points |
| Correct answer without justification or incorrect answer | 0 points |