Решить уравнение: [tex]sin^{2}\frac{x}{4}+sin^2\frac{3x}{4}=2cos^{2}\frac{x}{2}[/tex]. Created by ildarklopov. matematika-ru

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July 2022 1 17 Report

Решить уравнение:

[tex]sin^{2}\frac{x}{4}+sin^2\frac{3x}{4}=2cos^{2}\frac{x}{2}[/tex]

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$sin^2\frac{x}{4}+sin^2\frac{3x}{4}=2cos^2\frac{x}{2}\\\\zamena:\; \; a=\frac{x}{4}\; \; \Rightarrow \; \; \frac{3x}{4}=3a\; \; ,\; \; \frac{x}{2}=2a\; \; ,\; \; x=4a\\\\sin^2a+sin^23a=2cos^22a\\\\\frac{1-cos2a}{2}+\frac{1-cos6a}{2}=2\cdot \frac{1+cos4a}{2}\\\\1-cos2a+1-cos6a=2+2cos4a\\\\2-(cos2a+cos6a)=2+2cos4a\\\\-2\cdot cos4a\cdot cos2a-2\, cos4a=0\\\\-2\, cos4a\cdot (cos2a+1)=0\\\\a)\; \; cos4a=0\; \; \to \; \; 4a=\frac{\pi }{2}+\pi n\; ,\; n\in Z\\\\x=\frac{\pi}{2}+\pi n\; ,\; n\in Z$

$b)\; \; cos2a=-1\; \; \to \; \; 2a=\pi +2\pi k\; ,\; k\in Z\\\\\frac{x}{2}=\pi +2\pi k\; ,\; k\in Z\\\\x=2\pi +4\pi k\; ,\; k\in Z\\\\Otvet:\; \; x=\frac{\pi}{2}+\pi n,\; \; x=2\pi +4\pi k\; ,\; \; n,k\in Z\; .$

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