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Question Number 158774 by HongKing last updated on 08/Nov/21

Answered by MJS_new last updated on 08/Nov/21

sin^4  x +cos^2  x =sin^2  x +cos^4  x  ⇒ we have  2(√(7+cos 4x))+(√(9−cos 4x))=6(√2)  squaring & transforming 2 times  cos^2  4x −((242)/(25))cos 4x +((217)/(25))=0  cos 4x =((217)/(25)) [impossible]  cos 4x =1 ⇒ x=((nπ)/2)∧n∈Z

$$\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{2}} \:{x}\:=\mathrm{sin}^{\mathrm{2}} \:{x}\:+\mathrm{cos}^{\mathrm{4}} \:{x} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{2}\sqrt{\mathrm{7}+\mathrm{cos}\:\mathrm{4}{x}}+\sqrt{\mathrm{9}−\mathrm{cos}\:\mathrm{4}{x}}=\mathrm{6}\sqrt{\mathrm{2}} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming}\:\mathrm{2}\:\mathrm{times} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{4}{x}\:−\frac{\mathrm{242}}{\mathrm{25}}\mathrm{cos}\:\mathrm{4}{x}\:+\frac{\mathrm{217}}{\mathrm{25}}=\mathrm{0} \\ $$$$\mathrm{cos}\:\mathrm{4}{x}\:=\frac{\mathrm{217}}{\mathrm{25}}\:\left[\mathrm{impossible}\right] \\ $$$$\mathrm{cos}\:\mathrm{4}{x}\:=\mathrm{1}\:\Rightarrow\:{x}=\frac{{n}\pi}{\mathrm{2}}\wedge{n}\in\mathbb{Z} \\ $$

Commented by HongKing last updated on 09/Nov/21

perfect thank you so much my dear Ser

$$\mathrm{perfect}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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