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Question Number 143077 by mathdanisur last updated on 09/Jun/21

sin^5 x + cos^5 x = 2 − sin^4 x

$${sin}^{\mathrm{5}} {x}\:+\:{cos}^{\mathrm{5}} {x}\:=\:\mathrm{2}\:−\:{sin}^{\mathrm{4}} {x} \\ $$

Answered by MJS_new last updated on 10/Jun/21

sin x =s cos x =c cos^5 x =(1−sin^2 x)^2 cos x ⇒ s^5 +(1−s^2 )^2 c=2−s^4 s^5 +(c+1)s^4 −2cs^2 +c−2=0 (s−1)((s^3 +s^2 −s−1)c+(s^4 +2s^3 +2s^2 +2s+2))=0 s_1 =1 ⇒ sin x =1 ⇒ • x=(π/2)+2nπ • (s^3 +s^2 −s−1)c+(s^4 +2s^3 +2s^2 +2s+2)=0 c=f(s)=((s^4 +2s^3 +2s^2 +2s+2)/((1−s)(1+s)^2 )) −1≤s≤1 ⇒ f(s)≥2 ⇒ no other solution

$$\mathrm{sin}\:{x}\:={s} \\ $$$$\mathrm{cos}\:{x}\:={c} \\ $$$$\mathrm{cos}^{\mathrm{5}} \:{x}\:=\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} \mathrm{cos}\:{x} \\ $$$$\Rightarrow \\ $$$${s}^{\mathrm{5}} +\left(\mathrm{1}−{s}^{\mathrm{2}} \right)^{\mathrm{2}} {c}=\mathrm{2}−{s}^{\mathrm{4}} \\ $$$${s}^{\mathrm{5}} +\left({c}+\mathrm{1}\right){s}^{\mathrm{4}} −\mathrm{2}{cs}^{\mathrm{2}} +{c}−\mathrm{2}=\mathrm{0} \\ $$$$\left({s}−\mathrm{1}\right)\left(\left({s}^{\mathrm{3}} +{s}^{\mathrm{2}} −{s}−\mathrm{1}\right){c}+\left({s}^{\mathrm{4}} +\mathrm{2}{s}^{\mathrm{3}} +\mathrm{2}{s}^{\mathrm{2}} +\mathrm{2}{s}+\mathrm{2}\right)\right)=\mathrm{0} \\ $$$${s}_{\mathrm{1}} =\mathrm{1}\:\Rightarrow\:\mathrm{sin}\:{x}\:=\mathrm{1}\:\Rightarrow\:\bullet\:{x}=\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi\:\bullet \\ $$$$\left({s}^{\mathrm{3}} +{s}^{\mathrm{2}} −{s}−\mathrm{1}\right){c}+\left({s}^{\mathrm{4}} +\mathrm{2}{s}^{\mathrm{3}} +\mathrm{2}{s}^{\mathrm{2}} +\mathrm{2}{s}+\mathrm{2}\right)=\mathrm{0} \\ $$$${c}={f}\left({s}\right)=\frac{{s}^{\mathrm{4}} +\mathrm{2}{s}^{\mathrm{3}} +\mathrm{2}{s}^{\mathrm{2}} +\mathrm{2}{s}+\mathrm{2}}{\left(\mathrm{1}−{s}\right)\left(\mathrm{1}+{s}\right)^{\mathrm{2}} } \\ $$$$−\mathrm{1}\leqslant{s}\leqslant\mathrm{1}\:\Rightarrow\:{f}\left({s}\right)\geqslant\mathrm{2}\:\Rightarrow\:\mathrm{no}\:\mathrm{other}\:\mathrm{solution} \\ $$

Commented by mathdanisur last updated on 10/Jun/21

nice thanks sir

$${nice}\:{thanks}\:{sir} \\ $$

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