Consider $a=1,\:b\in \mathbb{R}$.
Show that there is single solution for the equation:$$x-a\sin x\:=\:b$$
So far I defined funtion $f\left(x\right)=x-a\sin \left(x\right)-b\:=\:x-\sin \left(x\right)-b$ and learned that $\lim _{x\to \infty }\left(f\left(x\right)\right)\:=\:\infty \:,\:\lim _{x\to -\infty }\left(f\left(x\right)\right)=-\infty \:$.
Now, the derivative of this function not always increasing because in $x_0=0$, $f'\left(x\right)\:=\:0$, so I can't use the Intermediate value theorem..
Any idea? tnx in advance!


Let $f(x)=x-\sin x-b$. Then $$f(1+b)=1-\sin(1+b)>0$$ and $$f(-1+b)=-1-\sin(-1+b)<0$$ but then since $f$ is continuous, it follows that $f$ has a root in $(-1+b,1+b)$. For the uniqueness of the root, suppose that $f$ has two roots $x_1$ and $x_2$. Then $$x_1 -x_2=\sin x_1 -\sin x_2.$$ Take absolute value of both sides and use $$\sin p -\sin q= 2\sin \frac{p-q}{2}\cos\frac{p+q}{2}$$ to get a contradiction!

Edit: since for any nonzero value of $x$ (in radian) one has $|\sin x|\lt |x|$ so

$|x_1 -x_2|=| \sin x_1 -\sin x_2|=2|\sin \frac{x_1-x_2}{2}||\cos\frac{x_1+x_2}{2}|<2\times |\frac{x_1-x_2}{2}|\times 1=|x_1 -x_2|$

  • $\begingroup$ tnx! but i'm little bit confused about what contradiction i need to see here.can you elaborate on this? $\endgroup$ – user2637293 Feb 19 '15 at 10:28
  • 1
    $\begingroup$ @user2637293 I edited the answer and added some details. $\endgroup$ – Fermat Feb 19 '15 at 14:33

Hints: use that $f_0(x)=x-\sin x$ is strictly increasing in $(-\infty,0)$ and $(0,+\infty)$. Also: $f_0<0$ in $(-\infty,0)$ and $f_0>0$ in $(0,+\infty)$


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