решите неравенства: cosx≤√3/2 sinx≥ -√2/2 sinx≤ 0 cosx≥ -√2/2 cosx≥ -√3/2 cosx< -√3/2 cosx≤ -1/2. Created by ponfilovden. algebra-ru

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July 2022 1 18 Report

решите неравенства:
cosx≤√3/2
sinx≥ -√2/2
sinx≤ 0
cosx≥ -√2/2
cosx≥ -√3/2
cosx< -√3/2
cosx≤ -1/2

Answers & Comments

F1staShka $cosx \leq \frac{ \sqrt{3} }{2} \\ \frac{ \pi }{6} +2 \pi n \leq x \leq \frac{11 \pi }{6} +2 \pi n$
$sinx \geq - \frac{ \sqrt{2} }{2} \\ - \frac{ \pi }{4} +2 \pi n \leq x \leq \frac{5 \pi }{4} +2 \pi n$
$sinx \leq 0 \\ \pi +2 \pi n \leq x \leq 2 \pi +2 \pi n$
$cosx \geq - \frac{ \sqrt{2} }{2} \\ - \frac{3 \pi }{4} +2 \pi n \leq x \leq \frac{3 \pi }{4} +2 \pi n$
$cosx \geq - \frac{ \sqrt{3} }{2} \\ - \frac{5 \pi }{6} +2 \pi n \leq x \leq \frac{5 \pi }{6}+2 \pi n$
$cosx\ \textless \ - \frac{ \sqrt{3} }{2} \\ \frac{5 \pi }{6}+2 \pi n\ \textless \ x\ \textless \ \frac{7 \pi }{6} +2 \pi n$
$cosx \leq - \frac{1}{2} \\ \frac{2 \pi }{3} +2 \pi n \leq x \leq \frac{4 \pi }{3} +2 \pi n$

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