It is given that:
[tex]\begin{gathered} 2\cos 2x+\sqrt[]{2}=0 \\ 2\cos 2x=-\sqrt[]{2} \\ \cos 2x=-\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]cos2x is negative square root 2 divided by 2 in the second and third quadrants, so it follows:
[tex]\begin{gathered} \cos 2x=\cos \theta \\ 2x=2n\pi\pm\theta \end{gathered}[/tex]Here theta is given by:
[tex]\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4},\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}[/tex]So the solution is given by:
[tex]\begin{gathered} 2x=2n\pi\pm\frac{3\pi}{4};2x=2n\pi\pm\frac{5\pi}{4} \\ x=n\pi\pm\frac{3\pi}{8};x=n\pi\pm\frac{5\pi}{8} \end{gathered}[/tex]