Помогите пожалуйста!!! [tex]cosx \geq - \frac{ \sqrt{2}}{2}[/tex] [tex] cosx \leq \frac{ \sqrt{3}}{2}[/tex]. Created by Olya3888. algebra-ru

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Olya3888 @Olya3888

July 2022 1 9 Report

Помогите пожалуйста!!!
[tex]cosx \geq - \frac{ \sqrt{2}}{2}[/tex]
[tex] cosx \leq \frac{ \sqrt{3}}{2}[/tex]

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TegeranUA

Verified answer

$cosx\geq-\frac{\sqrt{2}}{2} \\t_{_1}\leq x\leq t_{_1} \\t_{_1}=-arccos(-\frac{\sqrt2}{2})+2\pi n,\ n\in Z \\t_{_1}=-\pi+arccos(\frac{\sqrt2}{2})+2\pi n,\ n\in Z \\t_{_1}=-\pi+\frac{\pi}{4} +2\pi n,\ n\in Z \\t_{_1}=-\frac{3}{4}\pi+2\pi n,\ n\in Z \\t_{_2}=arccos(-\frac{\sqrt2}{2})+2\pi n,\ n\in Z \\t_{_2}=\frac{3}{4}\pi+2\pi n,\ n\in Z \\OTBET:x\in[-\frac{3}{4}\pi+2\pi n;\frac{3}{4}\pi+2\pi n],\ n\in Z$

$cosx\leq\frac{\sqrt3}{2}
\\t_{_1}\leq x \leq t_{_2}
\\t_{_1}=arccos(\frac{\sqrt3}{2})+2\pi n,\ n\in Z
\\t_{_1}=\frac{\pi}{6}+2\pi n,\ n\in Z
\\t_{_2}=2\pi-\frac{\pi}{6}+2\pi n,\ n\in Z
\\t_{_2}=\frac{11\pi}{6}+2\pi n,\ n\in Z
\\OTBET: x\in [\frac{\pi}{6}+2\pi n;\frac{11\pi}{6}+2\pi n],\ n\in Z$

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