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June 2022 1 38 Report

Решите неравенства:

[tex]|cosx|\geq \frac{\sqrt{3} }{2} \\cosx(2sin2x+\sqrt{3} )\ \textless \ 0[/tex]
[tex]\frac{\sqrt{3} }{3} \leq ctgx\leq4[/tex]

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NNNLLL54

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Ответ:

$1)\ \ |cosx|\geq \dfrac{\sqrt3}{2}\ \ \ \Rightarrow \ \ \ \left[\begin{array}{l}cosx\geq \dfrac{\sqrt3}{2}\\cos\leq -\dfrac{\sqrt3}{2}\end{array}\right\ \ \left[\begin{array}{l}x\in \Big[-\dfrac{\pi}{6}+2\pi n\ ;\ \dfrac{\pi}{6}+2\pi n\ \Big]\\x\in \Big[\ \dfrac{5\pi}{6}+2\pi n\ ;\ \dfrac{7\pi }{6}+2\pi n\ \Big]\end{array}\right\\\\\\x\in \Big[-\dfrac{\pi}{6}+2\pi n\ ;\ \dfrac{\pi}{6}+2\pi n\ \Big]\ \cup \ \Big[\ \dfrac{5\pi}{6}+2\pi n\ ;\ \dfrac{7\pi }{6}+2\pi n\ \Big]$

$2)\ \ cosx\cdot (2sin2x+\sqrt3)0\\sin2x0\\-\dfrac{2\pi }{3}+2\pi k$

$b)\ \ \left\{\begin{array}{l}cosx-\dfrac{\sqrt3}{2}\end{array}\right\ \ \left\{\begin{array}{l}cosx$

$c)\ \ x\in \Big(-\dfrac{\pi}{3} +2\pi k\ ;\ -\dfrac{\pi}{6}+2\pi k\ \Big)\cup \, \Big(\ \dfrac{5\pi}{6} +2\pi k\ ;\ \dfrac{3\pi}{2}+2\pi k\ \Big)\ ,\ k\in Z$

$3)\ \ \ \ \dfrac{\sqrt3}{3}\leq ctgx\leq 4\\\\x\in \Big[\ \dfrac{\pi}{3}+\pi n\ ;\ arcctg4+\pi n\ \Big]\ \ ,\ n\in Z$

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