Question

Решить неравенство:
0≤√2· cos(pi·x/4)<1

Answer 1

$0\leq \sqrt{2}\cos \cfrac{\pi x}{4} < 1\Leftrightarrow 0\leq \cos \cfrac{\pi x}{4} < \cfrac{1}{\sqrt{2}}\Leftrightarrow \begin{cases}\cos \cfrac{\pi x}{4}\geq 0\\ \cos \cfrac{\pi x}{4} < \cfrac{1}{\sqrt{2}}\end{cases}$$\begin{cases}\begin{cases}\cfrac{\pi x}{4}\geq -\cfrac{\pi}{2}+2\pi k\\ \cfrac{\pi x}{4}\leq \cfrac{\pi}{2}+2\pi k\end{cases}\\ \begin{cases}\cfrac{\pi x}{4} > \cfrac{\pi}{4}+2\pi k\\ \cfrac{\pi x}{4} < -\cfrac{\pi}{4}+2\pi k\end{cases}\end{cases},k\in \mathbb{Z}\Leftrightarrow \begin{cases}\begin{cases}x\geq -2+8k\\ x\leq 2+8k\end{cases}\\\begin{cases}x > 1+8k\\ x < -1+8k\end{cases}\end{cases}$$x\in [-2+8k,-1+8k)\cup (1+8k,2+8k],k\in \mathbb{Z}$