Найдите все значения при которых выполняется равенство f ' ( x )=0, если f ( x )=cos 2...

Решите задачу:

$f(x)=cos2x-x\sqrt2\; ,\; x\in [\, 0\, ;\, 2\pi \, ]\\\\f'(x)=-2\, sin2x-\sqrt2=0\\\\sin2x=-\dfrac{\sqrt2}{2}\; \; \; ,\; \; \; 2x=\left[\begin{array}{ccc}2x=-\frac{\pi}{4}+2\pi n\; ,\; n\in Z\\2x=-\frac{3\pi }{4}+2\pi k\; ,\; k\in Z\end{array}\right\\\\\\x=\left[\begin{array}{l}x=-\frac{\pi}{8}+\pi n\; ,\; n\in Z\\x=-\frac{3\pi}{8}+\pi k\; ,\; k\in Z\end{array}\right\\\\\\x\in [\, 0\, ;\, 2\pi \, ]:\; \; 0\leq -\frac{\pi}{8}+\pi n\leq 2\pi \; \; ,\; \; \; \; 0\leq -\frac{3\pi }{8}+\pi k\leq 2\pi$

$0\leq -\frac{1}{8}+n\leq 2\; \; \; ,\; \; \; 0\leq -\frac{3}{8}+k\leq 2\\\\\frac{1}{8}\leq n\leq 2\frac{1}{8}\; \; \; ,\; \; \; \frac{3}{8}\leq k\leq 2\frac{3}{8}\\\\n=1,2\; \; \; ,\; \; \; k=1,2\\\\n=1:\; \; x=-\frac{\pi}{8}+\pi =\frac{7\pi}{8}\\\\n=2:\; \; x=-\frac{\pi}{8}+2\pi =\frac{15\pi}{8}\\\\k=1:\; \; -\frac{3\pi}{8}+\pi =\frac{5\pi}{8}\\\\k=2:\; \; x=-\frac{3\pi}{8}+2\pi =\frac{13\pi }{8}\\\\Otvet:\; \; \frac{5\pi }{8}\; ,\; \frac{7\pi}{8}\; ,\; \frac{13\pi}{8}\; ,\; \frac{15\pi }{8}\; .$