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.
[Math] Prove that $\sin x – x\cos x = 0$ has only one solution in $ [-\frac{\pi}{2}, \frac{\pi}{2}]$
calculusrootstrigonometry
Best Answer
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.