$cos(2x+\dfrac{\pi}{4}=\dfrac{-1}{3}$
$⇔$\(\left[ \begin{array}{l}2x+\dfrac{\pi}{4}=arccos(\dfrac{-1}{3})+k2\pi\\2x+\dfrac{\pi}{4}=-arccos(\dfrac{-1}{3})+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}2x=-\dfrac{\pi}{4}+arccos(\dfrac{-1}{3})+k2\pi\\2x=-\dfrac{\pi}{4}-arccos(\dfrac{-1}{3})+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-\dfrac{\pi}{8}+\dfrac{1}{2}arccos(\dfrac{-1}{3})+k\pi\\x=-\dfrac{\pi}{8}-\dfrac{1}{2}arccos(\dfrac{-1}{3})+k\pi\end{array} \right.\)