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Question Number 120797 by Algoritm last updated on 02/Nov/20

Commented by liberty last updated on 02/Nov/20

$$\{\begin{array}{l}2\cos \mathrm{xcos}\mathrm{y}=3\sin \mathrm{y}\\ 2\cos \mathrm{ycos}\mathrm{z}=3\sin \mathrm{z}\\ 2\cos \mathrm{zcos}\mathrm{x}=3\sin \mathrm{x}\end{array}\to 8\cos {}^2\mathrm{xcos}{}^2\mathrm{ycos}{}^2\mathrm{z}=27\sin \mathrm{xsin}\mathrm{ysin}\mathrm{z}$$$$\mathrm{by}\mathrm{symetric}\{\begin{array}{l}2\cos {}^2\mathrm{x}=3\sin \mathrm{x}\\ 2\cos {}^2\mathrm{y}=3\sin \mathrm{y}\\ 2\cos {}^2\mathrm{z}=3\sin \mathrm{z}\end{array}$$$$\Rightarrow 2-2\sin {}^2\mathrm{x}=3\sin \mathrm{x};2\sin {}^2\mathrm{x}+3\sin \mathrm{x}-2=0$$$$(2\sin \mathrm{x}-1)(\sin \mathrm{x}+2)=0\Rightarrow \sin \mathrm{x}=\frac{1}{2}$$$$\mathrm{similarly}\mathrm{we}\mathrm{get}\to \{\begin{array}{l}\sin \mathrm{y}=\frac{1}{2}\\ \sin \mathrm{z}=\frac{1}{2}\end{array}$$$$\mathrm{checking}\{\begin{array}{l}\cos \mathrm{x}=\frac{\sqrt{3}}{2}\\ \tan \mathrm{y}=\frac{1}{\sqrt{3}}\end{array}\Rightarrow 2\cos \mathrm{x}=3\tan \mathrm{y}$$$$2\left(\frac{\sqrt{3}}{2}\right)=3\left(\frac{\sqrt{3}}{3}\right)\Rightarrow \sqrt{3}=\sqrt{3}\left(\mathrm{true}\right)$$

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