\({\log _2}\left( {2 + 2\sin \left( {x + y} \right)-{{\cos }^2}\left( {x + y} \right)} \right) = {4^x}-{2^{x + 1}} + 3\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,{\log _2}\left( {{{\sin }^2}\left( {x + y} \right) + 2\sin \left( {x + y} \right) + 1} \right) = {2^{2x}}-2 \cdot {2^x} + 3\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,{\log _2}{\left( {\sin \left( {x + y} \right) + 1} \right)^2} = {\left( {{2^x}-1} \right)^2} + 2.\)
Так как \(0 \le {\left( {\sin \left( {x + y} \right) + 1} \right)^2} \le 4,\) то левая часть \({\log _2}{\left( {\sin \left( {x + y} \right) + 1} \right)^2} \le 2\), а правая часть \({\left( {{2^x}-1} \right)^2} + 2 \ge 2\). Поэтому уравнение будет иметь решение, если:
\(\left\{ {\begin{array}{*{20}{c}}{{{\log }_2}{{\left( {\sin \left( {x + y} \right) + 1} \right)}^2} = 2,}\\{{{\left( {{2^x}-1} \right)}^2} + 2 = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\sin \left( {x + y} \right) = 1,\\x = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}y = \dfrac{{\rm{\pi }}}{2} + 2{\rm{\pi }}n,\\x = 0\end{array} \right.\,\,\,\,\,n \in Z.\)
Ответ: \(\left( {\,0;\,\dfrac{{\rm{\pi }}}{2} + 2{\rm{\pi }}\,n} \right),\;\;n \in Z.\)