Question Number 94297 by peter frank last updated on 17/May/20
Commented by mathmax by abdo last updated on 18/May/20
$${we}\:{know}\:{that}\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{x}={y}={z}\:\:\:\left({for}\:{x},{y}\:,{z}\:{reals}\right) \\ $$$${so}\:\left({e}\right)\:\Rightarrow{cosx}\:={cos}\left(\mathrm{2}{x}\right)={cos}\left(\mathrm{3}{x}\right)=\mathrm{0} \\ $$$${cosx}\:=\mathrm{0}\:\Rightarrow{x}\:=\frac{\pi}{\mathrm{2}}\:+{k}\pi\:\rightarrow{set}\:{w}_{\mathrm{1}{k}} \\ $$$${cos}\left(\mathrm{2}{x}\right)=\mathrm{0}\:\Rightarrow{x}\:=\frac{\pi}{\mathrm{4}}\:+\frac{{k}\pi}{\mathrm{2}}\rightarrow{set}\:{w}_{\mathrm{2}{k}} \\ $$$${cos}\left(\mathrm{3}{x}\right)=\mathrm{0}\:\Rightarrow{x}\:=\frac{\pi}{\mathrm{6}}\:+\frac{{k}\pi}{\mathrm{3}}\rightarrow{w}_{\mathrm{3}{k}} \\ $$$$\Rightarrow{x}\:\in\:\cap{w}_{{k}} \:\left({k}\in{Z}\right)\:{but}\:\frac{\pi}{\mathrm{2}}\:+{k}\pi\:=\frac{\pi}{\mathrm{4}}+{k}^{'} \pi\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}+{k}\:=\frac{\mathrm{1}}{\mathrm{4}}\:+{k}^{'} \:\Rightarrow \\ $$$$\mathrm{2}+\mathrm{4}{k}\:=\mathrm{1}\:+\mathrm{4}{k}^{'} \:\Rightarrow\mathrm{1}\:=\mathrm{4}\left({k}^{'} −{k}\right)\:\Rightarrow\mid{k}−{k}^{'} \mid=\frac{\mathrm{1}}{\mathrm{4}}\:\rightarrow{impossible} \\ $$$$\Rightarrow\:{x}\:{dont}\:{exist}\:\:{something}\:{went}\:{wrong}\:{in}\:{the}\:{Q}...! \\ $$