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Question Number 196523 by ERLY last updated on 26/Aug/23

resoudre dans c l equation sinx=2                                    <erly rolvinst>

$${resoudre}\:{dans}\:{c}\:{l}\:{equation}\:{sinx}=\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:<{erly}\:{rolvinst}> \\ $$

Answered by Frix last updated on 26/Aug/23

sin x =((e^(ix) −e^(−ix) )/(2i))=2  e^(ix) −(1/e^(ix) )=4i  e^(2ix) −4ie^(ix) −1=0  t=e^(ix)  ⇔ x=2nπ−i ln t  t^2 −4it−1=0  t=(2±(√3))i  ...  x=((4n+1)/2)π±i ln (2+(√3))

$$\mathrm{sin}\:{x}\:=\frac{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{e}^{−\mathrm{i}{x}} }{\mathrm{2i}}=\mathrm{2} \\ $$$$\mathrm{e}^{\mathrm{i}{x}} −\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{i}{x}} }=\mathrm{4i} \\ $$$$\mathrm{e}^{\mathrm{2i}{x}} −\mathrm{4ie}^{\mathrm{i}{x}} −\mathrm{1}=\mathrm{0} \\ $$$${t}=\mathrm{e}^{\mathrm{i}{x}} \:\Leftrightarrow\:{x}=\mathrm{2}{n}\pi−\mathrm{i}\:\mathrm{ln}\:{t} \\ $$$${t}^{\mathrm{2}} −\mathrm{4i}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right)\mathrm{i} \\ $$$$... \\ $$$${x}=\frac{\mathrm{4}{n}+\mathrm{1}}{\mathrm{2}}\pi\pm\mathrm{i}\:\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$

Answered by Skabetix last updated on 26/Aug/23

On pose z=x+iy avec x et y ∈ R  sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(y)  On obtient le systeme d equation suivant  sin(x)cosh(y)=2  (1)  cos(x)sinh(y)=0  (2)  En resolvant le systeme on a  x=(((pi)/2)+k×pi)(k pair)  y=±ln(2+(√)3)  il est possible de trouver toutes les solutions  en resolvant l equation  e^(it) −e^(−it) =4i

$${On}\:{pose}\:{z}={x}+{iy}\:{avec}\:{x}\:{et}\:{y}\:\in\:{R} \\ $$$${sin}\left({x}+{iy}\right)={sin}\left({x}\right){cosh}\left({y}\right)+{icos}\left({x}\right){sinh}\left({y}\right) \\ $$$${On}\:{obtient}\:{le}\:{systeme}\:{d}\:{equation}\:{suivant} \\ $$$${sin}\left({x}\right){cosh}\left({y}\right)=\mathrm{2}\:\:\left(\mathrm{1}\right) \\ $$$${cos}\left({x}\right){sinh}\left({y}\right)=\mathrm{0}\:\:\left(\mathrm{2}\right) \\ $$$${En}\:{resolvant}\:{le}\:{systeme}\:{on}\:{a} \\ $$$${x}=\left(\frac{{pi}}{\mathrm{2}}+{k}×{pi}\right)\left({k}\:{pair}\right) \\ $$$${y}=\pm{ln}\left(\mathrm{2}+\sqrt{}\mathrm{3}\right) \\ $$$${il}\:{est}\:{possible}\:{de}\:{trouver}\:{toutes}\:{les}\:{solutions} \\ $$$${en}\:{resolvant}\:{l}\:{equation} \\ $$$${e}^{{it}} −{e}^{−{it}} =\mathrm{4}{i} \\ $$

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