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Question For what $n\in [-1,1]$, the equation $\sin(x)=nx$ has (real) solutions other than $0$?

Attempt:

Following the suggestion from Tesla Daybreak, we consider $g(x)=\frac{\sin(x)}{x}, (1\mbox{ if }x=0)$ and find the range of $g$. We know $g(0)=1$, and \begin{align*} \sin(x)<x,\mbox{for all $x>0.$} \end{align*} Therefore, $1$ is the maximum value of $g(x)$ (Note $g$ is even).

For minimum point, we find $g'(x)$ to be $\frac{x\cos(x)-\sin(x)}{x^2}$. Setting this to be zero, we get for local minimum points $x_0$, $\tan(x_0)=x_0.$ Observe that

$$\tan(\pi)=0<\pi,\qquad \lim_{x\to \frac{\pi}{2}}\tan(x)=\infty>\frac{\pi}{2}.$$ $\tan(x)$ has a period of $\pi$. Therefore, $\tan(x)=x$ has solutions for $x\in (\pi,\frac{3\pi}{2})+k\pi$, for $k\in \mathbb{N_0}$. $g(x)=\frac{\sin(x)}{x}=\cos(x)\frac{\tan(x)}{x}$. For the minimum points $x_0$, $g(x_0)=\cos(x_0)<0$. The range of $g$ is $[\inf\{\cos(x_0)| \tan(x_0)=x_0\},1].$ This is the best I can come out with... I would appreciate a lot if someone can help to get better result.

123
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  • Consider using division instead of subtraction. – Tesla Daybreak Jul 29 '21 at 02:51
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    Is it not just all $n$ with real solution $x = 0$? – Ben Martin Jul 29 '21 at 03:11
  • @TeslaDaybreak Erm, but how? – 123 Jul 29 '21 at 03:13
  • @B.Martin Ahaha, I should exclude $0$ in the question. Thanks for the reminder. – 123 Jul 29 '21 at 03:13
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    @123 If you separate variables (i.e. make $x$ only appear on one side of equation and $n$ only on the other), then it simply becomes a problem of finding the range of a function in $x$. – Tesla Daybreak Jul 29 '21 at 03:26
  • @TeslaDaybreak AH, this is nice. Let me think about it. – 123 Jul 29 '21 at 03:28
  • We can also use graphs to plot the functions $y=nx$ and $y=\sin x$. This isn't very rigorous but I think it gives a general idea. For $n\in[0,1]$ the equation always has at least two solutions excluding $0$. – Aditya Jul 29 '21 at 04:02
  • For $-1<n<0$ we have to make the line $y=nx$ tangent to $y=\sin x$ ( it doesn't give two solutions for all $n\in[-1,0)$). – Aditya Jul 29 '21 at 04:05
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    @Aditya Ya. That is what I initially have done. But I would like to see more rigorous method. Anyway, thank you! – 123 Jul 29 '21 at 04:09
  • If the point of tangency is $(t,\sin t)$, $\frac{-3\pi}{2}<t<\frac{-\pi}{2}$ then $n=\frac{\sin t}{t}=\cos t \implies t=\tan t$. I couldn't solve this equation, but if someone can , then we have the complete range for $n$ such that $\sin x=nx$ has solutions other than $0$ as $n\in [\cos t, 1]$. – Aditya Jul 29 '21 at 04:09
  • Is $n$ necessarily natural? – Sayan Dutta Jul 29 '21 at 05:45
  • @SayanDutta NO. n is a real number between -1 and 1. – 123 Jul 29 '21 at 08:03
  • For the minimum I'm afraid an "exact" solution might not be possible, depending on your definition of exact. What you have is probably good enough. References: This question and this question. – Tesla Daybreak Jul 29 '21 at 16:42
  • @TeslaDaybreak I see. Seem like numerical approach is the best we can have – 123 Aug 01 '21 at 02:04

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