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Solve cos 2x = cos x for 0 <= x <= 2*pi.

 Feb 4, 2022
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Solve cos 2x = cos x for 0 <= x <= 2*pi.

 

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\(cos(2x)=cos(x)\\ 2cos^2(x)-1=cos(x)\\ cos(x)=z\\ 2z^2-z-1=0\ |\ /2\\ z^2-\frac{z}{2}-\frac{1}{2}=0\)

\(z=\frac{1}{4}\pm \sqrt{\frac{1}{16}+\frac{1}{2}}=\frac{1}{4}\pm \frac{3}{4}\\ cos(x)=z\\ cos(x)\in \{-\frac{1}{2},1\}\\ x\in \mathbb R\ |\ 0 \leq x\leq2\pi \\ {\color{blue}x}=arccos(-\frac{1}{2})\color{blue}\in \{\frac{2\pi}{3},\frac{4\pi}{3}\}\\ {\color{blue}x}=arccos(1)\color{blue}\in \{0,2\pi \}\)

laugh  !

 Feb 5, 2022
edited by asinus  Feb 5, 2022

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