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

sin^5 x+(√2)sinx=1        ,  x∈[0,2𝛑]

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

Answered by MJS last updated on 18/Dec/19

t=sin x  t^5 +(√2)tβˆ’1=0  only approximation possible  only one real solution  tβ‰ˆ.634429  β‡’ x=.687270∨x=2.45432

$${t}=\mathrm{sin}\:{x} \\ $$$${t}^{\mathrm{5}} +\sqrt{\mathrm{2}}{t}βˆ’\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{only}\:\mathrm{approximation}\:\mathrm{possible} \\ $$$$\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{solution} \\ $$$${t}\approx.\mathrm{634429} \\ $$$$\Rightarrow\:{x}=.\mathrm{687270}\vee{x}=\mathrm{2}.\mathrm{45432} \\ $$

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 βˆ’20sin^2  x +5)sin x  sin x =t  16t^5 βˆ’20t^3 +(5+(√2))t=1  t^5 βˆ’(5/4)t^3 +((5+(√2))/(16))tβˆ’(1/(16))=0  approximation  t_1 β‰ˆ.171177  t_2 β‰ˆ.551260  t_3 β‰ˆ.929365  t_(4, 5) β‰ˆβˆ’.825901Β±.174831i  β‡’  x_(11) β‰ˆ.172024; x_(12) β‰ˆ2.96957  x_(21) β‰ˆ.583873; x_(22) β‰ˆ2.55772  x_(31) β‰ˆ1.19269; x_(32) β‰ˆ1.94890

$$\mathrm{sin}\:\mathrm{5}{x}\:+\sqrt{\mathrm{2}}\mathrm{sin}\:{x}\:=\mathrm{1} \\ $$$$\mathrm{sin}\:\mathrm{5}{x}\:=\left(\mathrm{16sin}^{\mathrm{4}} \:{x}\:βˆ’\mathrm{20sin}^{\mathrm{2}} \:{x}\:+\mathrm{5}\right)\mathrm{sin}\:{x} \\ $$$$\mathrm{sin}\:{x}\:={t} \\ $$$$\mathrm{16}{t}^{\mathrm{5}} βˆ’\mathrm{20}{t}^{\mathrm{3}} +\left(\mathrm{5}+\sqrt{\mathrm{2}}\right){t}=\mathrm{1} \\ $$$${t}^{\mathrm{5}} βˆ’\frac{\mathrm{5}}{\mathrm{4}}{t}^{\mathrm{3}} +\frac{\mathrm{5}+\sqrt{\mathrm{2}}}{\mathrm{16}}{t}βˆ’\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0} \\ $$$$\mathrm{approximation} \\ $$$${t}_{\mathrm{1}} \approx.\mathrm{171177} \\ $$$${t}_{\mathrm{2}} \approx.\mathrm{551260} \\ $$$${t}_{\mathrm{3}} \approx.\mathrm{929365} \\ $$$${t}_{\mathrm{4},\:\mathrm{5}} \approxβˆ’.\mathrm{825901}\pm.\mathrm{174831i} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{11}} \approx.\mathrm{172024};\:{x}_{\mathrm{12}} \approx\mathrm{2}.\mathrm{96957} \\ $$$${x}_{\mathrm{21}} \approx.\mathrm{583873};\:{x}_{\mathrm{22}} \approx\mathrm{2}.\mathrm{55772} \\ $$$${x}_{\mathrm{31}} \approx\mathrm{1}.\mathrm{19269};\:{x}_{\mathrm{32}} \approx\mathrm{1}.\mathrm{94890} \\ $$

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