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](https://www.tinkutara.com/question/Q158785.png)
$$\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{2}} \:{x}\:=\mathrm{sin}^{\mathrm{2}} \:{x}\:+\mathrm{cos}^{\mathrm{4}} \:{x} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{2}\sqrt{\mathrm{7}+\mathrm{cos}\:\mathrm{4}{x}}+\sqrt{\mathrm{9}−\mathrm{cos}\:\mathrm{4}{x}}=\mathrm{6}\sqrt{\mathrm{2}} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming}\:\mathrm{2}\:\mathrm{times} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{4}{x}\:−\frac{\mathrm{242}}{\mathrm{25}}\mathrm{cos}\:\mathrm{4}{x}\:+\frac{\mathrm{217}}{\mathrm{25}}=\mathrm{0} \\ $$$$\mathrm{cos}\:\mathrm{4}{x}\:=\frac{\mathrm{217}}{\mathrm{25}}\:\left[\mathrm{impossible}\right] \\ $$$$\mathrm{cos}\:\mathrm{4}{x}\:=\mathrm{1}\:\Rightarrow\:{x}=\frac{{n}\pi}{\mathrm{2}}\wedge{n}\in\mathbb{Z} \\ $$
Commented by HongKing last updated on 09/Nov/21
$$\mathrm{perfect}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$