sin-5-x-2-sinx-1-x-0-2-

Question Number 75846 by behi83417@gmail.com last updated on 18/Dec/19

$$\boldsymbol{\mathrm{sin}}^{\mathrm{5}} \boldsymbol{\mathrm{x}}+\sqrt{\mathrm{2}}\boldsymbol{\mathrm{sinx}}=\mathrm{1}\:\:\:\:\:\:\:\:,\:\:\boldsymbol{\mathrm{x}}\in\left[\mathrm{0},\mathrm{2}\boldsymbol{\pi}\right] \\ $$

Answered by MJS last updated on 18/Dec/19

t=sin x t^5 +(√2)t−1=0 only approximation possible only one real solution t≈.634429 ⇒ x=.687270∨x=2.45432

$${t}=\mathrm{sin}\:{x} \\ $$$${t}^{\mathrm{5}} +\sqrt{\mathrm{2}}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{only}\:\mathrm{approximation}\:\mathrm{possible} \\ $$$$\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{solution} \\ $$$${t}\approx.\mathrm{634429} \\ $$$$\Rightarrow\:{x}=.\mathrm{687270}\vee{x}=\mathrm{2}.\mathrm{45432} \\ $$

Commented by behi83417@gmail.com last updated on 18/Dec/19

thank you proph:MJS. this is my typo.the original question is: sin5x+(√2)sinx=1.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{proph}:\mathrm{MJS}. \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{my}\:\mathrm{typo}.\mathrm{the}\:\mathrm{original}\:\mathrm{question}\:\mathrm{is}: \\ $$$$\boldsymbol{\mathrm{sin}}\mathrm{5}\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{2}}\boldsymbol{\mathrm{sinx}}=\mathrm{1}. \\ $$

Answered by MJS last updated on 18/Dec/19

sin 5x +(√2)sin x =1 sin 5x =(16sin^4 x −20sin^2 x +5)sin x sin x =t 16t^5 −20t^3 +(5+(√2))t=1 t^5 −(5/4)t^3 +((5+(√2))/(16))t−(1/(16))=0 approximation t_1 ≈.171177 t_2 ≈.551260 t_3 ≈.929365 t_(4, 5) ≈−.825901±.174831i ⇒ x_(11) ≈.172024; x_(12) ≈2.96957 x_(21) ≈.583873; x_(22) ≈2.55772 x_(31) ≈1.19269; x_(32) ≈1.94890

$$\mathrm{sin}\:\mathrm{5}{x}\:+\sqrt{\mathrm{2}}\mathrm{sin}\:{x}\:=\mathrm{1} \\ $$$$\mathrm{sin}\:\mathrm{5}{x}\:=\left(\mathrm{16sin}^{\mathrm{4}} \:{x}\:−\mathrm{20sin}^{\mathrm{2}} \:{x}\:+\mathrm{5}\right)\mathrm{sin}\:{x} \\ $$$$\mathrm{sin}\:{x}\:={t} \\ $$$$\mathrm{16}{t}^{\mathrm{5}} −\mathrm{20}{t}^{\mathrm{3}} +\left(\mathrm{5}+\sqrt{\mathrm{2}}\right){t}=\mathrm{1} \\ $$$${t}^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{4}}{t}^{\mathrm{3}} +\frac{\mathrm{5}+\sqrt{\mathrm{2}}}{\mathrm{16}}{t}−\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0} \\ $$$$\mathrm{approximation} \\ $$$${t}_{\mathrm{1}} \approx.\mathrm{171177} \\ $$$${t}_{\mathrm{2}} \approx.\mathrm{551260} \\ $$$${t}_{\mathrm{3}} \approx.\mathrm{929365} \\ $$$${t}_{\mathrm{4},\:\mathrm{5}} \approx−.\mathrm{825901}\pm.\mathrm{174831i} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{11}} \approx.\mathrm{172024};\:{x}_{\mathrm{12}} \approx\mathrm{2}.\mathrm{96957} \\ $$$${x}_{\mathrm{21}} \approx.\mathrm{583873};\:{x}_{\mathrm{22}} \approx\mathrm{2}.\mathrm{55772} \\ $$$${x}_{\mathrm{31}} \approx\mathrm{1}.\mathrm{19269};\:{x}_{\mathrm{32}} \approx\mathrm{1}.\mathrm{94890} \\ $$

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