# Problem of the Week #77 - September 15th, 2013

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#### Chris L T521

##### Well-known member
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Show that the equation $2x-1-\sin x=0$ has exactly one real solution.

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Hint:
First use IVT to show that $2x-1-\sin x=0$ has at least one real solution. Then use Rolle's theorem to show that you can't have two or more solutions.

#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by anemone, eddybob123, johng, and MarkFL. You can find anemone's solution below.

Let $f(x)=2x-1-\sin x$ and we see that $f(x)$ is a continuous function for all real $x$.

We're asked to show that the equation $f(x)=2x-1-\sin x=0$ has exactly one real solution. One way to prove that is to first show that the function of $f$ of $x$ has a root and then show that the root is unique.

First, we suspect there is a root lies beween $$\displaystyle \frac{5 \pi}{18}$$ and $$\displaystyle \frac{\pi}{30}$$. Plugging in both values into $f(x)$, we get $f(\frac{5 \pi}{18}) \approx -0.02$ and $f(\frac{\pi}{30}) \approx 0.22$. Since $f(\frac{5 \pi}{18})<0$ and $f(\frac{\pi}{30})>0$ and $f(x)$ is a continuous function for all real $x$, by the Intermediate Value Theorem, there is a number $c$ between $$\displaystyle \frac{5 \pi}{18}$$ and $$\displaystyle \frac{\pi}{30}$$ such that $f'(c)=0$. Thus, $f(x)$ must have at least a real solution.

Next, we must show that this root is unique. Our plan is to prove it by contradiction. By the definition of Rolle's theorem, if we first assume the equation $2x-1-\sin x=0$ has at least two roots a and b, that is, $f(a)=0$ and $f(b)=0$, and then there is at least one point d in $(a, b)$ where $f'(d)=0$.

$f(x)=2x-1-\sin x$

$f'(x)=2-\cos x$

$f'(d)=0$ iff $2-\cos d=0$, or $\cos d =2$ But this is impossible since $-1<\cos d<1$.

This gives a contradiction and therefore, the equation can't have two roots, and the only root is unique.

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