Помогите пожалуйста решить уравнение [tex]2sin {}^{2} x - \sqrt{3} sin2x = 0[/tex]. Created by olesechka2004. algebra-ru

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olesechka2004 @olesechka2004

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Помогите пожалуйста решить уравнение
[tex]2sin {}^{2} x - \sqrt{3} sin2x = 0[/tex]

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$\displaystyle 2sin^2x-\sqrt{3}sin2x=0\\\\2sin^2x-2\sqrt{3}sinx*cosx=0\\\\2sinx(sinx-\sqrt{3}cosx)=0\\\\\\\\sinx=0; x=\pi n; n \in Z\\\\sinx-\sqrt{3}cosx=0 |:cosx\\\\ tgx-\sqrt{3}=0; tgx=\sqrt{3}; x=\frac{\pi }{3}+\pi n; n \in Z$

ответ:

$\displaystyle \pi n; n \in Z\\\\\frac{\pi }{3}+\pi n; n \in Z$

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olesechka2004 August 2022 | 0 Ответы

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izvestnochto n naturalnoe chislokakim chislomchetnym ili nechetnymyavlyaetsya znache

teplovoj dvigatel poluchil ot nagrevatelya 6 kdzhi otdal holodilniku kolichestvo

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