3tgx+√3>0 Решите фастом. Created by mrdesantnik21. algebra-ru

3tgx+√3>0 Решите фастом...

July 2022 1 8 Report

3tgx+√3>0
Решите фастом

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red321 $3tgx+\sqrt{3}>0\\tgx>-\frac{\sqrt{3}}{3}\\x\in(arctg(-\frac{\sqrt{3}}{3})+\pi*n;\frac{\pi}{2}+\pi*n),\ n\in Z\\x\in(-\frac{\pi}{6}+\pi*n;\frac{\pi}{2}+\pi*n),\ n\in Z$

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