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Let $\alpha\geq 0$, prove that $\alpha \sin x+x\cos x=0$ if and only if $x=\frac{\pi}{2}+n\pi +\varepsilon_n$ for some $n$ with $\varepsilon_n\to \frac{\alpha}{\pi}$ when $n\to\infty$

It is clear that if $\alpha=0$ then $x\cos x=0$ and thus $x=0$ or $x=\frac{\pi}{2}+n\pi$ for some integer $n\geq 0$

If $\alpha\ne 0$ what can I do?

FUUNK1000
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    What is the connection between $x$ and $x_n$ ??? –  Jun 20 '16 at 18:52
  • Is there any reason for my edit not being accepted ? –  Jun 20 '16 at 20:20
  • Write $\alpha \sin x+x\cos x$ in the form $\sqrt{\alpha^2 + x^2} \sin (x + \beta)$, with $\tan \beta = \frac{x}{\alpha}$. Since the square root is never zero, the zeroes are determined by $x + \beta = n \pi$, i.e. $x + \tan^{-1} \frac{x}{\alpha} = n \pi$. – jim Jun 20 '16 at 22:20

1 Answers1

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Continuing from Jim work, we will work on solution

$$x+tan^{-1} \frac x{\alpha}=n\pi$$

transform this to finding x of the followings, $$\frac x{\alpha}=\tan (n\pi-x)= -\tan x$$

For solution, you can find answer in here

the linear function $y=-\frac x{\alpha}$ has many intersections with $y=\tan x$. One intersection $(x_n, y_n)$ in interval between $(n \pi -\frac {\pi}2, n \pi +\frac {\pi}2)$. Or $x_n = n \pi - \frac {\pi}2 + \epsilon_n$. Or because it is a period function, $x_n = n \pi + \frac {\pi}2 + \epsilon_n$

Replacing the root,

$$-\frac {n \pi + \frac {\pi}2 + \epsilon_n}{\alpha}= \tan (n \pi + \frac {\pi}2 + \epsilon_n)=\tan (\frac {\pi}2 + \epsilon_n)=-\frac 1{\tan \epsilon _n}$$

Thus, $$\frac {\alpha}{ \pi + \frac {\frac {\pi}2 + \epsilon_n}n}= n\tan \epsilon _n$$

when $n \rightarrow \infty$, $\epsilon_n \rightarrow \frac{\alpha}2$

This is a little different from the posted result $\frac{\alpha}{\pi}$

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user115350
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