\frac1{\sqrt2}\\ x=\frac{\pi}3\Rightarrow\cos x=\frac12<\frac1{\sqrt2}\\ x=-\frac{\pi}6\Rightarrow\cos x=\frac{\sqrt3}2>\frac1{\sqrt2}\\ x\in[-\frac{\pi}4+2\pi n;\frac{\pi}4+2\pi n],\quad n\in\mathbb{Z}" alt="\cos x \geq\frac1{\sqrt2}\\ -1\leq\cos x\leq1\\ \cos x=\frac1{\sqrt2}\\ x=\pm\frac{\pi}4+2\pi n\\ x=\frac{\pi}6\Rightarrow\cos x=\frac{\sqrt3}2>\frac1{\sqrt2}\\ x=\frac{\pi}3\Rightarrow\cos x=\frac12<\frac1{\sqrt2}\\ x=-\frac{\pi}6\Rightarrow\cos x=\frac{\sqrt3}2>\frac1{\sqrt2}\\ x\in[-\frac{\pi}4+2\pi n;\frac{\pi}4+2\pi n],\quad n\in\mathbb{Z}" align="absmiddle" class="latex-formula">