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Question Number 61162 by Tawa1 last updated on 29/May/19
if sin(x) = ((x − 20)/(20)) , find x
$$\mathrm{if}\:\:\:\:\:\mathrm{sin}\left(\mathrm{x}\right)\:\:=\:\:\frac{\mathrm{x}\:−\:\mathrm{20}}{\mathrm{20}}\:\:,\:\:\:\mathrm{find}\:\:\mathrm{x} \\ $$
Commented by kaivan.ahmadi last updated on 29/May/19
we can find number of solution by plote y=sinx and y=((x−20)/(20)) this equation has 13 answer
$${we}\:{can}\:{find}\:{number}\:{of}\:{solution}\:{by}\:{plote} \\ $$$${y}={sinx}\:{and}\:{y}=\frac{{x}−\mathrm{20}}{\mathrm{20}} \\ $$$${this}\:{equation}\:{has}\:\mathrm{13}\:{answer} \\ $$
Commented by maxmathsup by imad last updated on 29/May/19
(e) ⇔sin(x) =(x/(20)) −1 changement (x/(20)) =t give sin(20t) =t−1 ⇔ sin(20t)−t +1 =0 let f(t) =sin(20t)−t+1 ⇒f^′ (t) =20cos(t)−1 f^′ (t) =0 ⇒20cost =1 ⇔cost =(1/(20)) let α /cos(α)=(1/(20)) ⇒ t =α +2kπ or t =−α +2kπ ⇒t =arcos((1/(20)))+2kπ or t =−arcos((1/(20))) +2kπ in this case it better to use newton method to approximate the roots....
$$\left({e}\right)\:\Leftrightarrow{sin}\left({x}\right)\:=\frac{{x}}{\mathrm{20}}\:−\mathrm{1}\:\:{changement}\:\frac{{x}}{\mathrm{20}}\:={t}\:{give} \\ $$$${sin}\left(\mathrm{20}{t}\right)\:={t}−\mathrm{1}\:\Leftrightarrow\:{sin}\left(\mathrm{20}{t}\right)−{t}\:+\mathrm{1}\:=\mathrm{0} \\ $$$${let}\:{f}\left({t}\right)\:={sin}\left(\mathrm{20}{t}\right)−{t}+\mathrm{1}\:\Rightarrow{f}^{'} \left({t}\right)\:=\mathrm{20}{cos}\left({t}\right)−\mathrm{1} \\ $$$${f}^{'} \left({t}\right)\:=\mathrm{0}\:\Rightarrow\mathrm{20}{cost}\:=\mathrm{1}\:\Leftrightarrow{cost}\:=\frac{\mathrm{1}}{\mathrm{20}}\:{let}\:\alpha\:/{cos}\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{20}}\:\Rightarrow \\ $$$${t}\:=\alpha\:+\mathrm{2}{k}\pi\:\:{or}\:{t}\:=−\alpha\:+\mathrm{2}{k}\pi\:\Rightarrow{t}\:={arcos}\left(\frac{\mathrm{1}}{\mathrm{20}}\right)+\mathrm{2}{k}\pi\:{or}\: \\ $$$${t}\:=−{arcos}\left(\frac{\mathrm{1}}{\mathrm{20}}\right)\:+\mathrm{2}{k}\pi\:\:\:{in}\:{this}\:{case}\:{it}\:{better}\:{to}\:{use}\:\:{newton} \\ $$$${method}\:\:{to}\:{approximate}\:{the}\:{roots}…. \\ $$
Commented by Tawa1 last updated on 30/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by malwaan last updated on 29/May/19
−1≤sin(x)≤1 ⇒−1≤((x−20)/(20))≤1 ⇒−20≤x−20≤20 ⇒0≤ x≤40 ⇒x∈[ 0 ; 40]
$$−\mathrm{1}\leqslant{sin}\left({x}\right)\leqslant\mathrm{1} \\ $$$$\Rightarrow−\mathrm{1}\leqslant\frac{{x}−\mathrm{20}}{\mathrm{20}}\leqslant\mathrm{1} \\ $$$$\Rightarrow−\mathrm{20}\leqslant{x}−\mathrm{20}\leqslant\mathrm{20} \\ $$$$\Rightarrow\mathrm{0}\leqslant\:\boldsymbol{{x}}\leqslant\mathrm{40} \\ $$$$\Rightarrow\boldsymbol{{x}}\in\left[\:\mathrm{0}\:;\:\mathrm{40}\right] \\ $$
Commented by Tawa1 last updated on 30/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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