Question Number 190297 by Shrinava last updated on 31/Mar/23
Answered by Frix last updated on 31/Mar/23
$$\mathrm{Let}\:{u}=\mathrm{cos}\:{x}\:\wedge{v}=\mathrm{sin}\:{x} \\ $$$$\mathrm{Transforming}\:\mathrm{leads}\:\mathrm{to} \\ $$$${u}^{\mathrm{6}} +\mathrm{2}{u}^{\mathrm{3}} {v}^{\mathrm{3}} +\mathrm{2}{u}^{\mathrm{2}} {v}^{\mathrm{2}} −\mathrm{2}{uv}+{v}^{\mathrm{6}} =\mathrm{0} \\ $$$${v}=\sqrt{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$${u}^{\mathrm{4}} −{u}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{u}\left({u}^{\mathrm{4}} −{u}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left({u}^{\mathrm{4}} −{u}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−\mathrm{2}{u}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{only}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{is}\:{u}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Leftrightarrow \\ $$$$\mathrm{cos}\:{x}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow \\ $$$${x}=\pm\frac{\pi}{\mathrm{4}}+\mathrm{2}{n}\pi;\:{n}\in\mathbb{Z} \\ $$