Prove that there exists only one number $x \in \mathbb{R}$ such that $\sin(x)=x-1$.

Question

Prove that there exists only one number $x \in \mathbb{R}$ such that $\sin(x)=x-1$.

I have been looking after this problem now on Holidays, but I can't prove the uniqueness.

Let $f(x)=\sin(x)$ and $g(x)=x-1$. Both functions are continuous everywhere, particularly in the interval $[1,2]$. We also may say that both functions are bounded in that interval by $0\leq f(x) \leq 1$ and $0\leq g(x) \leq 1$ (computing), so that we can say that there exists a number $x$ such that $\sin(x)=x-1$.

But now I can't prove that $x$ is unique.

Answer 1

Let $$f(x) = \sin x -x+1 $$

Note that $$f(0)=1 >0$$

and $$ f(\pi )=-\pi +1 <0$$

The intermediate value theorem implies that there is some $x\in (0,\pi ) $ for which $f(x)=0.$

Note that $$f'(x)=cos(x)-1 \le 0$$ which implies that your function is non-increasing and there are no turning points.

Thus the solution is unique.

Answer 2

The question is whether the function $f(x)\colon x\mapsto 1 + \sin x$ has at most one fixed point. You can check that $|f(x_1)-f(x_2)|<|x_1-x_2|$ for $x_1\ne x_2$.

ADDED: Indeed $$f(x_1)-f(x_2)= \sin x_1- \sin x_2= 2 \sin \frac{x_1-x_2}{2}\cdot \cos\frac{x_1+x_2}{2}$$ Now use the fact that $|\sin x| <|x|$ for $x\ne 0$ (real).

Answer 3

The extrema of $f(x):=\sin x-x+1$ are solutions of

$$\cos x=1$$

or $$x_k=2k\pi.$$

Plugging these values,

$$f(x_k)=\sin2k\pi-2k\pi+1=-2k\pi+1,$$ there is a single change of sign, between $k=0$ and $k=1$.