Proof $x=\sin(x+1)$ has one solution in $\mathbb{R}$ I have this problem : 
Proof $x=\sin(x+1)$ has one solution in $\mathbb{R}$.
I got stuck and I don't go how to "move" on.
My proof
$f(x)=\sin(x+1)-x$
If $f$ is injective function then $x=\sin(x+1)$ has one solution.
$f'(x)=\cos(x+1)-1$, but $f'(x)$ is not increasing/decreasing since $x=2 \pi-1 \rightarrow f'(x)=0$, so $f$ is not injective function.
I don't have any idea how to approach this.
Any ideas? Any help will be appreciated.
 A: Draw a picture of the two functions: $x$, $\sin(x-1)$. Identify an interval where the difference is monotone and changes sign. There you have one and only one solution. Prove that outside such interval $|x|>1$ while $|\sin(x-1)|\le 1$ so there are not other solutions.
A: There is an easily identified, very large class of possible values for $x$ that cannot possibly be a solution. If you identify that class, then you can restrict further analysis to the remaining numbers; e.g. maybe you can show that $f$, restricted to that interval, is injective!
A: Let $y=x+1$; then the equation becomes 
$$
\sin y = y-1$$
Lemma 1: $\sin y = y-1$ has no solutions (in $\Bbb{R}$) with $y < 0$.
Proof: $\forall y: |\sin y| \leq 1$  so $\forall y < 0: \sin y \geq -|\sin y| \geq -1 
> -1+y = y-1$ so $\sin y > y-1$ and the equality cannot hold.
Lemma 2: $y = 0$ is not a solution to $\sin y = y-1$.  Indeed, $\sin 0 = 0 \neq -1$.
Lemma 3: $\sin y = y-1$ has no solutions  with $y > \pi$.
Proof:  $\forall y: \sin y \leq 1$  so $\forall y \geq \pi : \sin y \leq 1 
< \pi-1 \leq  y-1$ so $\sin y < y-1$ and the equality cannot hold.
Lemma 4:  $\sin y = y-1$ has no solutions  with $0 < y \leq  \pi/2$.
Proof: At $y = \pi/2$, $\sin y - (y-1) > 0$.  In the interval $(0,\pi/2]$, 
$\frac{d}{dy} (\sin y - (y-1)) = \cos y -1 < 0$.  Therefore, $\sin y - (y-1) > 0$ is a monotonic decreasing continuous function with positive endpoint value on this interval, therefore it must be positive on the whole interval.
Lemma 5: All solutions to  $\sin y = y-1$ lie in $(\pi/2,\pi)$.  Obtained by combining lemma's 1 through 4.
Lemma 6:  $\sin y = y-1=0$ has at most one solution in $(\pi/2,\pi)$. 
Proof:  In $(\pi/2,\pi)$, $\cos y <0$ so  $\frac{d}{dy} (\sin y - (y-1)) = \cos y -1 < 0$.  So we have a monotonic decreasing continuous function in the interval, which can have at most one zero.
Lemma 7:  $\sin y = y-1=0$ has at least one solution in $(\pi/2,\pi)$. 
Proof: $g(y) \equiv (\sin y - (y-1))$ is continuous.  $g(\pi/2) = 2 - \pi/2 > 0$.
$g(\pi) = 1-\pi < 0$.  By the intermediate value theorem, there must be some $y$ in the interval such that $\sin y = y-1=0$ .
Lemma 8: $\sin y = y-1=0$ has exactly  one solution in $(\pi/2,\pi)$. (By combining lemmas 6 and 7.)
Proposition: $\sin y = y-1$ has exactly one real solution.  (Combine lemmas 5 and 8.)
Therefore, $\sin (x+1) = x$ has exactly one real solution. 
A: Use your function $f(x)=\sin(x+1)-x$.
Note that $f(0)>0$ and $f(2)<0$. Since $f$ is continuous, there is at least one zero in the interval $[0,2]$.
Next note that $f'(x) \le 0$ for all $x$, with equality happening only at isolated points $x=2k\pi-1$ for $k \in \mathbb Z$. This means that $f$ is strictly decreasing and therefore injective. Therefore there is at most one zero in the real numbers.
A: Let $g(x) = x - \sin(x+1)$, we have
$$g'(x) = 1 - \cos(x+1) \ge 0\quad\text{ for all } x \in \mathbb{R}$$
This implies $g(x)$ is non-decreasing. In fact, $g(x)$ is strictly increasing! 
To see this, we use the fact 
$$g'(x) = 0 \quad\text{ when and only when }\quad x = (\ell + \frac12)\pi - 1 \quad\text{ for }\quad \ell \in \mathbb{Z}$$
This means the zeroes of $g'(x)$ are isolated. Let's say $a, b$ are two successive zeros of $g'(x)$. For any two numbers $x, y$ such that $a < x < y < b$, we can use MVT to find three
number $\mu, \nu, \omega$ such that 
$$a < \mu < x < \nu < y < \omega < b\quad\text{ and }\quad 
\begin{cases}
g(b) - g(y) &= g'(\omega) (b - y) > 0\\
g(y) - g(x) &= g'(\nu) (y - x) > 0\\
g(x) - g(a) &= g'(\mu) (x - a) > 0
\end{cases}$$
From this, we can conclude $g(x)$ is strictly increasing on $[a,b]$ and hence over $\mathbb{R}$.
Since $g(x)$ is strictly increasing, it is injective.
Notice $g(x)$ is unbounded from above as $x \to \infty$ and unbounded from below as $x \to -\infty$.
Using IVT, we can conclude $g(x)$ is surjective.
Combine this, we find $g(x)$ is bijective and hence for any $\lambda \in \mathbb{R}$, the equation
$$g(x) = \lambda \quad\iff\quad x = \sin(x+1) + \lambda$$
has a unique solution. In particular, you can fix $\lambda$ to $0$ and deduce the equation
$$x = \sin(x+1)$$
has a unique solution.
A: You don't care if $f'(x)$ is increasing or decreasing, you care if $f(x)$ is.  In particular, if you can show that it is monotone increasing or monotone decreasing then you know that it has at most one zero, ie the orignal equation has at most one solution.  What property of $f'(x)$ tells you whether $f(x)$ is increasing or decreasing?
A: It's easier to see what's going on if you let $x+1=u$, so that the equation becomes $u-1=\sin u$.  So what you want to show is that the sine curve and a certain simple line intersect just once.  If you draw the two curves, you can see they intersect somewhere in the range $0\lt u\lt \pi$.  You can prove there's at least one intersection by looking at the endpoints and noting that $-1\lt\sin0$ while $\pi-1\gt\sin\pi$.  Furthermore, you can easily show that there are no intersections for $u\lt0$ or $u\gt\pi$, because $|u-1|\gt1$ in each of those ranges.
Finally, to show there's only one intersection for $0\lt u\lt\pi$, note that
$$f(u)=u-1-\sin u\implies f'(u)=1-\cos u\gt0$$
in this range.
Strictly speaking, it's not necessary to switch from $x$ to $u$.  But if you don't, then you have to draw a shifted sine curve in order to see what's going on, and I always shift things the wrong way.
A: You're right that your proof doesn't show that $f$ is injective, but it doesn't show that it's not either - which is good, because $f$ actually is injective. Your proof suffices to show that $f$ is non-increasing, however, which we  can use to show injectivity:
Suppose that there were two points, $a,b$ such that $f(a)=f(b)$. By $f$ being non-increasing, we can  conclude that for any $c\in[a,b]$, it holds that $f(c)=f(a)=f(b)$. Therefore, $f'(c)=0$. This is impossible to be satisfied in any interval where $a\neq b$ because the set of points where $f'(x)=0$ is the countable set $2\pi n - 1$ for $n\in \mathbb{Z}$. Therefore, if $f(a)=f(b)$ then $a=b$.
