Prove that $\sin x - x\cos x = 0$ has only one solution in $ [-\frac{\pi}{2}, \frac{\pi}{2}]$

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I have to prove that $\sin x - x\cos x = 0$ has only one solution in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. While it seems obvious that one solution might be $x=0$, I don't know how to do a formal proof.

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2 Answers

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Let $f(x)=\sin x-x\cos x$. You have $f'(x)=x\sin x$. Since $\sin x$ has the same sign as $x$ for $x\in[-\pi/2,\pi/2]$, we know that $f'(x)\geq0$ in this interval and $f'(x)>0$ for $x\in[-\pi/2,\pi/2]\setminus\{0\}$. Therefore $f$ is strictly increasing and can have at most one zero.

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$$\sin x - x\cos x = 0 \iff \sin x = x\cos x.$$ Divide through by $\cos x$ to get $$\tan x = x.$$

So you've just reduced the problem to showing that $\tan x = x$ (both monotone functions in the interval) has one solution. The normal analysis-y way of proving this is true is in another answer, so I thought I'd do a pretty-picture method in mine.

A quick sketch of the $\tan x$ function reveals that the line $y=x$ only cuts through $\tan x$ once, at the origin. (In the relevant interval)

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