\frac{\sqrt2}2\\ x\in[-\frac{\pi}4+2\pi n,\frac{\pi}4+2\pi n],\quad n\in\mathbb{Z}" alt="2\cos x+\sqrt2\leq0\\ \cos x\leq-\frac{\sqrt2}2\\ \cos x=\frac{\sqrt2}2\\ x=\pm\frac{\pi}4+2\pi n\\ x=\pm\frac{\pi}2\Rightarrow\cos x=0<\frac{\sqrt2}2\\ x=0\Rightarrow\cos x=1>\frac{\sqrt2}2\\ x\in[-\frac{\pi}4+2\pi n,\frac{\pi}4+2\pi n],\quad n\in\mathbb{Z}" align="absmiddle" class="latex-formula">