найдите нули функции y=2sin(2x-pi/3). Created by nika5956. algebra-ru

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August 2022 1 5 Report

найдите нули функции y=2sin(2x-pi/3)

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genius20

$y=2 \sin (2x-\pi /3)\\\\2\sin (2x-\pi/3)=0\\\sin (2x-\pi/3)=0\\\\2x-\dfrac{\pi}{3}=\pi k, \quad k \in \mathbb Z\\\\2x= \dfrac{\pi}{3}+\pi k, \quad k \in \mathbb Z \\\\x=\dfrac{\pi}{6}+\dfrac{\pi k }{2}, \quad k \in \mathbb Z$

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