\(\sqrt 2 \,\cos \left( {\dfrac{\pi }{4} + 2x} \right)-\cos 2x = \dfrac{1}{2}\)
Воспользуемся формулой косинуса суммы:
\(\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta -\sin \alpha \sin \beta .\)
\(\sqrt 2 \left( {\cos \dfrac{\pi }{4}\cos 2x-\sin \dfrac{\pi }{4}\sin 2x} \right)-\cos 2x = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\sqrt 2 \cdot \dfrac{{\sqrt 2 }}{2}\cos 2x-\sqrt 2 \cdot \dfrac{{\sqrt 2 }}{2}\sin 2x-\cos 2x = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\cos 2x-\sin 2x-\cos 2x = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-\sin 2x = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\)
\( \Leftrightarrow \,\,\,\,\,\,\,\sin 2x = -\dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x = -\dfrac{\pi }{6} + 2\pi k,}\\{2x = -\dfrac{{5\pi }}{6} + 2\pi k}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -\dfrac{\pi }{{12}} + \pi k,}\\{x = -\dfrac{{5\pi }}{{12}} + \pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(-\dfrac{\pi }{{12}} + \pi k;\;\;\;-\dfrac{{5\pi }}{{12}} + \pi k;\;\;\;k \in Z.\)