\(\left| {\cos x + \sin x} \right| = \sqrt 2 \sin 2x\)
Так как левая часть уравнения неотрицательна (за счёт модуля), то для того, чтобы уравнение имело решение, правая часть также должна быть неотрицательной: \(\sqrt 2 \sin 2x \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\sin 2x \ge 0.\)
Тогда:
\(\left| {\cos x + \sin x} \right| = \sqrt 2 \sin 2x\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\sin 2x \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {\cos x + \sin x} \right)}^2} = {{\left( {\sqrt 2 \sin 2x} \right)}^2}}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\sin 2x \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\cos }^2}x + 2\cos x\sin x + {{\sin }^2}x = 2{{\sin }^2}2x}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.\left\{ {\begin{array}{*{20}{c}}{\sin 2x \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2{{\sin }^2}2x-\sin 2x-1 = 0}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\sin 2x \ge 0,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{\sin 2x = 1,\,\,\,\,\,}\\{\sin 2x = -\dfrac{1}{2}}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sin 2x = 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = \dfrac{\pi }{4} + \pi k,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(\dfrac{\pi }{4} + \pi k,\,\,\,k\, \in \,Z.\)