Ответ:

[tex]\left [ \begin{array} {l} \dfrac{\pi }{2} +\pi n \\\\ (-1)^{n+1}\cdot \dfrac{\pi }{12}+ \dfrac{\pi n}{2}~~ , ~ n \in \mathbb {Z} \end{array}[/tex]

Объяснение:

[tex]\displaystyle \sin 2x - \cos 2x= 1+\sqrt{2} \cos x \\\\[/tex]

Воспользуемся формулами

[tex]\sf cos ~ 2a = cos^2 a -sin^2a \\\\ sin~2a= 2~ sin ~a\cdot cos ~a[/tex]

Тогда

[tex]\displaystyle \sin 2x - \cos 2x= 1+\sqrt{2} \cos x\\\\ 2\sin x \cos x - (\cos ^2x- \sin^2x)=\sin ^2x+\cos ^2x +\sqrt{2} \cos x \\\\ 2\sin x\cos x-\cos^2x+\sin ^2x =\sin^2+\cos^2x+\sqrt{2} \cos x \\\\ 2\cos ^2x+\sqrt{2}\cos x-2\sin x\cos x =0 \\\\ 2\cos x\bigg (\cos x-\sin x + \frac{\sqrt{2} }{2} ~ \bigg ) =0[/tex]

Каждую скобку приравняем к нулю

[tex]1) ~~ \cos x = 0 \Leftrightarrow\boxed{ x = \frac{\pi }{2} +\pi n~~ , ~ n \in \mathbb Z}[/tex]

[tex]\displaystyle \hspace{-1em} 2) ~ \cos x -\sin x+\dfrac{\sqrt{2} }{2} =0 \\\\ (\cos x-\sin x)^2 =\bigg(-\cfrac{\sqrt{2} }{2} ~ \bigg )^2 \\\\ \sin ^2x+\cos ^2 x -2 \sin x\cos x = \cfrac{1}{2} \\\\ 1-\sin 2x =\cfrac{1}{2} \\\\ \sin 2x =\frac{1}{2} \\\\ 2x =(-1)^{n+1}\arcsin \frac{1}{2} +\pi n \\\\ 2x =(-1)^{n+1} \cdot \cfrac{\pi }{6} +\pi n \\\\ \boxed{x =(-1)^{n+1}\cdot \frac{\pi }{12}+ \frac{\pi n}{2}~~ , ~ n \in \mathbb Z }[/tex]