\section*{Problem 5 - 051045}

Determine all real numbers \(x\) and \(y\) that satisfy the equation:

\[
[\sin (x-y)+1] \cdot[2 \cos (2 x-y)+1]=6
\]
}

Due to \(-1 \leq \sin \alpha \leq 1\) and \(-1 \leq \cos \alpha \leq 1\), it holds that

\[
[\sin (x-y)+1] \cdot[2 \cos (2 x-y)+1] \leq 6
\]

for all real \(x\) and \(y\), and the equality sign stands precisely when

\[
\sin (x-y)=1 \quad ; \quad \cos (2 x-y)=1
\]

holds simultaneously. The system of equations (1) is therefore equivalent to the given equation and is satisfied precisely when there exist two integers \(m\) and \(n\) such that

\[
x-y=\frac{\pi}{2}+2 m \pi \quad ; \quad 2 x-y=2 n \pi
\]

holds. Therefore, all solutions to the given equation are obtained when \(k=n-m\) and \(m\) in the equations

\[
\begin{gathered}
x=-\frac{\pi}{2}+2(n-m) \pi=-\frac{\pi}{2}+2 k \pi \\
y=[2(n-2 m)-1] \pi=[2(k-m)-1] \pi
\end{gathered}
\]

independently run through all integers.

Adapted from [5]