სავარჯიშო 3.6.3 Mar 5, 2022 16:05:39 GMT 4 Quote Select PostDeselect PostLink to PostMemberGive GiftBack to Top Post by Evgeny on Mar 5, 2022 16:05:39 GMT 4 იპოვეთ \[f^{-1}\Bigg(\bigcup^2_{n=1}A_n\Bigg)\] სადაც $A_1 = [0; \pi]$, $A_2 = [\pi; 2\pi]$, ხოლო $f:\mathbb{R}\to \mathbb{R}$ ფუნქცია განმარტებულია ასე: $f(x) = \sin x$
სავარჯიშო 3.6.3 Mar 21, 2022 3:10:14 GMT 4 Quote Select PostDeselect PostLink to PostMemberGive GiftBack to Top Post by taraxa on Mar 21, 2022 3:10:14 GMT 4 $$\bigcup\limits^{2}_{n=1}A_{n} = [0; 2\pi]$$$$f^{-1}([0; 2\pi]) = \{ x | f(x) \in [0; 2\pi] \} = \bigcup\limits^{\infty}_{k=0} [\pm2\pi k; \pi \pm 2\pi k]$$
