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Question Number 190297 by Shrinava last updated on 31/Mar/23

Answered by Frix last updated on 31/Mar/23

$$\mathrm{Let}u=\cos x\wedge v=\sin x$$$$\mathrm{Transforming}\mathrm{leads}\mathrm{to}$$$$u^6+2u^3{v}^3+2u^2{v}^2-2uv+v^6=0$$$$v=\sqrt{1-u^2}$$$$u^4-u^2+1-2u(u^4-u^2+1)\sqrt{1-u^2}=0$$$$(u^4-u^2+1)(1-2u\sqrt{1-u^2})=0$$$$\mathrm{The}\mathrm{only}\mathrm{real}\mathrm{solution}\mathrm{is}u=\frac{\sqrt{2}}{2}\iff$$$$\cos x=\frac{\sqrt{2}}{2}\Rightarrow$$$$x=\pm \frac{\pi }{4}+2n\pi ;n\in \mathbb{Z}$$

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