Function with only one real root

Question

I'm trying to show that the function $f(x)=2x+3sinx+xcosx$ has only one real root (which is $0$) I've noticed that this is an odd function and therefore if it has a second real root $x_0>0$ then $-x_0$ will be a root as well. Then I tried using Rolle's Theorem but that didn't get me anywhere. Any hints?

Answer 1

Focusing on $\mathbb{R}_+$ by parity ($f$ is odd). Using the fact that $$-x-3 \leq 3\sin x + x\cos x \leq x+3$$ you get $ f(x)\geq x-3$ for all $x\geq 0$. Therefore, for $x>3$, $f(x)$ cannot be zero. Furthermore, for $x\in [0,3]$ you have $\sin x \geq 0$, so with the same technique the inequality now becomes $f(x) \geq x$, and $f$ cannot be zero on $(0,3]$ either.

Answer 2

Hint: the magnitude of the second term is at most 3, the magnitude of the third term is at most $|x|$.