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Question Number 210702 by depressiveshrek last updated on 17/Aug/24
How many real solutions does the equation x=sin3x have?
$$\mathrm{How}\:\mathrm{many}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{does}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}=\mathrm{sin3}{x}\:\mathrm{have}? \\ $$
Answered by Frix last updated on 17/Aug/24
x=sin 3x t=3x (t/3)=sin t t_1 =0 −1≤(t/3)≤1 −3≤t≤3 sin t ≥0 ⇒ 0≤t≤π; π>3 ⇒ 1 solution for t∈(0, 3] 1 solution for t∈[−3, 0) ⇒ 3 solutions (t/n)=sin t −n≤t≤n sin t ≥0 ⇒ 2kπ≤t≤(2k+1)π First time more than 3 solutions possible at n=7 −7≤t≤7 0≤t≤π∨2π≤t≤3π but in fact (t/7)>sin t; 2π≤t n=8 ⇒ (t/8)=sin t has 7 solutions
$${x}=\mathrm{sin}\:\mathrm{3}{x} \\ $$$${t}=\mathrm{3}{x} \\ $$$$\frac{{t}}{\mathrm{3}}=\mathrm{sin}\:{t} \\ $$$${t}_{\mathrm{1}} =\mathrm{0} \\ $$$$−\mathrm{1}\leqslant\frac{{t}}{\mathrm{3}}\leqslant\mathrm{1} \\ $$$$−\mathrm{3}\leqslant{t}\leqslant\mathrm{3} \\ $$$$\mathrm{sin}\:{t}\:\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{0}\leqslant{t}\leqslant\pi;\:\pi>\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{1}\:\mathrm{solution}\:\mathrm{for}\:{t}\in\left(\mathrm{0},\:\mathrm{3}\right] \\ $$$$\:\:\:\:\:\:\mathrm{1}\:\mathrm{solution}\:\mathrm{for}\:{t}\in\left[−\mathrm{3},\:\mathrm{0}\right) \\ $$$$\Rightarrow\:\mathrm{3}\:\mathrm{solutions} \\ $$$$ \\ $$$$ \\ $$$$\frac{{t}}{{n}}=\mathrm{sin}\:{t} \\ $$$$−{n}\leqslant{t}\leqslant{n} \\ $$$$\mathrm{sin}\:{t}\:\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{2}{k}\pi\leqslant{t}\leqslant\left(\mathrm{2}{k}+\mathrm{1}\right)\pi \\ $$$$\mathrm{First}\:\mathrm{time}\:\mathrm{more}\:\mathrm{than}\:\mathrm{3}\:\mathrm{solutions}\:\mathrm{possible} \\ $$$$\mathrm{at}\:{n}=\mathrm{7} \\ $$$$−\mathrm{7}\leqslant{t}\leqslant\mathrm{7} \\ $$$$\mathrm{0}\leqslant{t}\leqslant\pi\vee\mathrm{2}\pi\leqslant{t}\leqslant\mathrm{3}\pi \\ $$$$\mathrm{but}\:\mathrm{in}\:\mathrm{fact}\:\frac{{t}}{\mathrm{7}}>\mathrm{sin}\:{t};\:\mathrm{2}\pi\leqslant{t} \\ $$$$ \\ $$$${n}=\mathrm{8}\:\Rightarrow\:\frac{{t}}{\mathrm{8}}=\mathrm{sin}\:{t}\:\mathrm{has}\:\mathrm{7}\:\mathrm{solutions} \\ $$
Answered by mr W last updated on 17/Aug/24
say 3x=t, the equation becomes (t/3)=sin t. let′s generall look at the equation kx=sin x. we see x=0 is always a root. besides due to symmetry if x=a is a root, then x=−a is also a root. therefore the number of roots is always odd. due to symmetry we only need to treat k≥0. for x=0, sin x=0 has infinite roots. so now we take k>0.
$${say}\:\mathrm{3}{x}={t},\:{the}\:{equation}\:{becomes}\: \\ $$$$\frac{{t}}{\mathrm{3}}=\mathrm{sin}\:{t}.\:{let}'{s}\:{generall}\:{look}\:{at}\:{the} \\ $$$${equation}\:{kx}=\mathrm{sin}\:{x}.\: \\ $$$${we}\:{see}\:{x}=\mathrm{0}\:{is}\:{always}\:{a}\:{root}.\:{besides} \\ $$$${due}\:{to}\:{symmetry}\:{if}\:{x}={a}\:{is}\:{a}\:{root},\: \\ $$$${then}\:{x}=−{a}\:{is}\:{also}\:{a}\:{root}.\:{therefore} \\ $$$${the}\:{number}\:{of}\:{roots}\:{is}\:{always}\:{odd}. \\ $$$${due}\:{to}\:{symmetry}\:{we}\:{only}\:{need}\:{to} \\ $$$${treat}\:{k}\geqslant\mathrm{0}.\:{for}\:{x}=\mathrm{0},\:\mathrm{sin}\:{x}=\mathrm{0}\:{has} \\ $$$${infinite}\:{roots}.\:{so}\:{now}\:{we}\:{take}\:{k}>\mathrm{0}. \\ $$
Commented by mr W last updated on 17/Aug/24
Commented by mr W last updated on 17/Aug/24
(kx)′=(sin x)′ ⇒k=cos x=(√(1−k^2 x^2 )) ⇒x=(√((1/k^2 )−1)) 2(n−1)π<x<(2(n−1)+(1/2))π 4(n−1)^2 π^2 <x^2 <4(n−(3/4))^2 π^2 4(n−1)^2 π^2 +1<(1/k_n ^2 )<4(n−(3/4))^2 π^2 +1 (1/( (√(4(n−(3/4))^2 π^2 +1))))<k_n <(1/( (√(4(n−1)^2 π^2 +1)))) k_n =cos (√((1/k_n ^2 )−1)) this equation has infinite solutions: k_1 , k_2 , k_3 , ... k_1 =1 k_2 ≈(1/(7.789706)) k_3 ≈(1/(14.101695)) k_4 ≈(1/(20.395833)) ...... at x=0: k=k_1 =1. if k≥k_1 =1, there is one root x=0. if k<k_1 =1, there are at least 3 roots. generally for k=k_n there are 4n−3 roots for k_(n+1) <k<k_n there are 4n−1 roots back to the original question with x=sin 3x. it is the same as (x/3)=sin x since k=(1/3)<k_1 , >k_2 ⇒it has 3 roots.
$$\left({kx}\right)'=\left(\mathrm{sin}\:{x}\right)'\: \\ $$$$\Rightarrow{k}=\mathrm{cos}\:{x}=\sqrt{\mathrm{1}−{k}^{\mathrm{2}} {x}^{\mathrm{2}} }\:\Rightarrow{x}=\sqrt{\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\mathrm{1}} \\ $$$$\mathrm{2}\left({n}−\mathrm{1}\right)\pi<{x}<\left(\mathrm{2}\left({n}−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)\pi \\ $$$$\mathrm{4}\left({n}−\mathrm{1}\right)^{\mathrm{2}} \pi^{\mathrm{2}} <{x}^{\mathrm{2}} <\mathrm{4}\left({n}−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \pi^{\mathrm{2}} \\ $$$$\mathrm{4}\left({n}−\mathrm{1}\right)^{\mathrm{2}} \pi^{\mathrm{2}} +\mathrm{1}<\frac{\mathrm{1}}{{k}_{{n}} ^{\mathrm{2}} }<\mathrm{4}\left({n}−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \pi^{\mathrm{2}} +\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\left({n}−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \pi^{\mathrm{2}} +\mathrm{1}}}<{k}_{{n}} <\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\left({n}−\mathrm{1}\right)^{\mathrm{2}} \pi^{\mathrm{2}} +\mathrm{1}}} \\ $$$${k}_{{n}} =\mathrm{cos}\:\sqrt{\frac{\mathrm{1}}{{k}_{{n}} ^{\mathrm{2}} }−\mathrm{1}} \\ $$$${this}\:{equation}\:{has}\:{infinite}\:{solutions}: \\ $$$${k}_{\mathrm{1}} ,\:{k}_{\mathrm{2}} ,\:{k}_{\mathrm{3}} ,\:… \\ $$$${k}_{\mathrm{1}} =\mathrm{1} \\ $$$${k}_{\mathrm{2}} \approx\frac{\mathrm{1}}{\mathrm{7}.\mathrm{789706}} \\ $$$${k}_{\mathrm{3}} \approx\frac{\mathrm{1}}{\mathrm{14}.\mathrm{101695}} \\ $$$${k}_{\mathrm{4}} \approx\frac{\mathrm{1}}{\mathrm{20}.\mathrm{395833}} \\ $$$$…… \\ $$$${at}\:{x}=\mathrm{0}:\:{k}={k}_{\mathrm{1}} =\mathrm{1}.\: \\ $$$${if}\:{k}\geqslant{k}_{\mathrm{1}} =\mathrm{1},\:{there}\:{is}\:{one}\:{root}\:{x}=\mathrm{0}. \\ $$$${if}\:{k}<{k}_{\mathrm{1}} =\mathrm{1},\:{there}\:{are}\:{at}\:{least}\:\mathrm{3}\:{roots}. \\ $$$$ \\ $$$${generally} \\ $$$${for}\:{k}={k}_{{n}} \:{there}\:{are}\:\mathrm{4}{n}−\mathrm{3}\:{roots} \\ $$$${for}\:{k}_{{n}+\mathrm{1}} <{k}<{k}_{{n}} \:{there}\:{are}\:\mathrm{4}{n}−\mathrm{1}\:{roots} \\ $$$$ \\ $$$${back}\:{to}\:{the}\:{original}\:{question}\:{with} \\ $$$${x}=\mathrm{sin}\:\mathrm{3}{x}. \\ $$$${it}\:{is}\:{the}\:{same}\:{as}\:\frac{{x}}{\mathrm{3}}=\mathrm{sin}\:{x} \\ $$$${since}\:{k}=\frac{\mathrm{1}}{\mathrm{3}}<{k}_{\mathrm{1}} ,\:>{k}_{\mathrm{2}} \:\Rightarrow{it}\:{has}\:\mathrm{3}\:{roots}. \\ $$

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