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I am trying to show that there is only one solution for the equation $x-\cos x = 1$ in the interval $]0,\frac{\pi}{2}[$.

Using $f(x) = x-\cos x -1 = 0$, I took the derivative $1 + \sin x$.

Now I would expect to find solutions for $1 + \sin x = 0$ within the interval, but the next candidate to the left is $-\frac{\pi}{2}$, which is outside of the interval - this does not prove the existence of a solution within the interval, does it?.

What am I doing wrong?

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  • $\begingroup$ In the interval $(0,\pi/2)$ the derivative is positive, so the function is… Next, $\lim_{x\to0}f(x)=-1$ and $\lim_{x\to\pi/2}f(x)=(\pi-2)/2$, so… $\endgroup$ – egreg Nov 5 '16 at 21:03
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Rolles theorem does not apply here : For Rolles theorem you need two distinct real numbers $a$ and $b$ with $\ f(a)=f(b)\ $. Here, no such pair within the interval $\ [0,\frac{\pi}{2}]\ $ exists.

The correct way is using $\ f'(x)=1+\sin(x)>0\ $ to show that there is at most one solution and looking at the signs of $f(0)$ and $f(\frac{\pi}{2})$ to see that there is at least one solution.

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Roots of the derivative are possible locations of minima or maxima.

That there is no root shows that $f$ is monotonous over the interval.

Now check the function values at the interval ends.


As $\cos x \simeq 1-\frac12x^2$ the roots are not too far from the solutions of the quadratic equation $x-2+\frac{x^2}2=0$ $\iff$ $(x+1)^2=3$ $\iff$ $x=-1\pm \sqrt3$, i.e., $x=0.73205...$ for the root in the interval.

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Let $f(x)=x-\cos(x)-1$ . we use the reasoning absurd and we assume there are two roots $a$ and $\; b$ in $(0,\frac{\pi}{2})$ such that

$f(a)=f(b)=0.$

$f$ is continuous at $[a,b]$ and diffetentiable at $(a,b)$, thus using Rolle's theorem, there will exist $c\in(a,b)\subset (0,\frac{\pi}{2})$ such that

$f'(c)=1+\sin(c)=0$ which is in contradiction with $\;\;\sin(c)>0$. So, there is only one root in $(0,\frac{\pi}{2})$.

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    $\begingroup$ You can only conclude that there is at most one root. There is one because of the intermediate value theorem. $\endgroup$ – egreg Nov 5 '16 at 22:21
  • $\begingroup$ @egreg I supposed there are two and this gives contradiction . $\endgroup$ – hamam_Abdallah Nov 6 '16 at 14:13

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