tex]...

July 2022 2 42 Report

30 баллов. Знатоки алгебры, помогите, пожалуйста, решить уравнение.
[tex] \sqrt{3} \cos(2x) + \sin(2x) = \sqrt{2} [/tex]

NNNLLL54

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$\sqrt3cos2x+sin2x=\sqrt2\ \Big|:2\\\\\\\dfrac{\sqrt3}{2}\cdot cos2x+\dfrac{1}{2}\cdot sin2x=\dfrac{\sqrt2}{2}\\\\\\sin\dfrac{\pi}{3}\cdot cos2x+cos\dfrac{\pi}{3}\cdot sin2x=\dfrac{\sqrt2}{2}\\\\\\sin\Big(2x+\dfrac{\pi}{3}\Big)=\dfrac{\sqrt2}{2}\\\\\\2x+\dfrac{\pi}{3}=(-1)^{n}\cdot \dfrac{\pi}{4}+\pi n\ ,\ n\in Z\\\\\\2x=(-1)^{n}\cdot \dfrac{\pi}{4}-\dfrac{\pi}{3}+\pi n\ ,\ n\in Z\\\\\\x=(-1)^{n}\cdot \dfrac{\pi}{8}-\dfrac{\pi}{6}+\dfrac{\pi n}{2}\ ,\ n\in Z$

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ata221 спасибо!!!

sharofat0

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Ответ:

x(1)=-pi/24+pin ; n€Z

x(2)=5/24pi+pin ; n€Z.

1 votes Thanks 1

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