11.3. Solve the system of equations $\left\{\begin{array}{l}\sin x \cos y=\sin z \\ \cos x \sin y=\cos z\end{array}\right.$, if the numbers $x, y$ and $z$ lie in the interval $\left[0 ; \frac{\pi}{2}\right]$
Answer: $\left(\frac{\pi}{2} ; 0 ; \frac{\pi}{2}\right),\left(0 ; \frac{\pi}{2} ; 0\right)$

Solution. First method. Square both equations. We get: $\left\{\begin{array}{l}\sin ^{2} x \cos ^{2} y=\sin ^{2} z \\ \cos ^{2} x \sin ^{2} y=\cos ^{2} z\end{array}\right.$

Add the equations:

$\sin ^{2} x \cos ^{2} y+\cos ^{2} x \sin ^{2} y=1 \Leftrightarrow$

$\sin ^{2} x \cos ^{2} y+\cos ^{2} x \sin ^{2} y=\sin ^{2} x+\cos ^{2} x$.

Since $\sin ^{2} x \cos ^{2} y \leq \sin ^{2} x$ and $\cos ^{2} x \sin ^{2} y \leq \cos ^{2} x$, the obtained equality can only hold when equality is achieved in each of the used inequalities.

Thus, we need to consider four cases:

1) $\cos ^{2} y=1$ and $\sin ^{2} y=1$, which is impossible, as it contradicts the fundamental trigonometric identity.
2) $\cos ^{2} y=0$ and $\sin ^{2} y=0$, which is also impossible for the same reasons.
3) $\cos ^{2} y=1$ and $\sin ^{2} y=0$, then $\sin y=0, \cos z=0, \sin ^{2} x=1$, i.e., $\cos x=0$. Considering that the values of all variables lie in the interval $\left[0 ; \frac{\pi}{2}\right]$, we get: $x=\frac{\pi}{2} ; y=0 ; z=\frac{\pi}{2}$.
4) $\cos ^{2} y=0$ and $\sin ^{2} y=1$, then $\cos y=0, \sin z=0, \cos ^{2} x=1$, i.e., $\sin x=0$. Considering that the values of all variables lie in the interval $\left[0 ; \frac{\pi}{2}\right]$, we get: $x=0 ; y=\frac{\pi}{2} ; z=0$.

Substituting the found solutions into the original system, we verify that they satisfy the condition.

Second method. Adding and subtracting the equations of the system term by term, we obtain an equivalent system: $\left\{\begin{array}{l}\sin (x+y)=\sin z+\cos z, \\ \sin (x-y)=\sin z-\cos z\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin (x+y)=\sqrt{2} \cos \left(z-\frac{\pi}{4}\right), \\ \sin (x-y)=\sqrt{2} \sin \left(z-\frac{\pi}{4}\right)\end{array}\right.\right.$. Squaring each equation and adding them, and considering the fundamental trigonometric identity, we get the consequence: $\sin ^{2}(x+y)+\sin ^{2}(x-y)=2$. Since $\sin ^{2}(x+y) \leq 1$ and $\sin ^{2}(x-y) \leq 1$, we have $\left\{\begin{array}{l}\sin ^{2}(x+y)=1, \\ \sin ^{2}(x-y)=1\end{array}\right.$.

Since $0 \leq x \leq \frac{\pi}{2}$ and $0 \leq y \leq \frac{\pi}{2}$, then $0 \leq x+y \leq \pi$ and $-\frac{\pi}{2} \leq x-y \leq \frac{\pi}{2}$. Considering these inequalities, we get: $\left\{\begin{array}{l}x+y=\frac{\pi}{2}, \\ x-y= \pm \frac{\pi}{2}\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{\pi}{2}, \\ y=0\end{array}\right.\right.$ or $\left\{\begin{array}{l}x=0, \\ y=\frac{\pi}{2}\end{array}\right.$. Substituting the obtained values into the original system, we get that $z=\frac{\pi}{2}$ or $z=0$ respectively.

## Evaluation Criteria.

 “+” A complete and well-reasoned solution is provided
 “士” A generally correct reasoning is provided, containing minor gaps or inaccuracies (for example, in the first method, the verification is missing)
 “Ғ” The correct answer is provided and verified to satisfy the condition
 “Ғ” A correct idea of the solution is present, but not fully developed or developed with errors

 “-” Only the answer is provided

 “-” The problem is not solved or solved incorrectly