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