How to demonstrate that $x + \sin(x) = 0$ has a single root?

Question

I have the following expression:

$$x + \sin(x)=0 $$

I have to demonstrate that this expression has a single root.

What can I do

Answer 1

The derivate of $f(x)=x+\sin(x)$ is $f'(x)=1+\cos(x)$ which is non-negative and $0$ only for isolated points.

Hence, $f(x)$ is strictly increasing, therefore it can have at most one root.

Since $f(0)=0$, $0$ is the only root.

Answer 2

Clearly $0$ is a root.

For $0x>0$, and for $x\ge \pi$, we have $x+\sin x\ge x-1\ge \pi-1>0$. Hence there is no positive root. As $x\mapsto x+\sin x$ is odd, there is also no negative root.

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Remark: This uses very little about the sine function, not even continuity.

Answer 3

Peter's answer is perfect.

And for the sake of curiosity, if you are searching for a root near the origin, for example, then you can use Taylor expansion for the sine:

$$\sin(x) \approx x - \frac{x^3}{6} + \mathcal{O}(x^3)$$

hence

$$x - x + \frac{x^3}{6} = 0$$

$$x^3 = 0$$

Which means $x = 0$. Three solutions, but identical.