- Thread starter
- Moderator
- #1

- Jan 26, 2012

- 995

-----

-----

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

-----

**Hint**:

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter Chris L T521
- Start date

- Status
- Not open for further replies.

- Thread starter
- Moderator
- #1

- Jan 26, 2012

- 995

-----

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter
- Moderator
- #2

- Jan 26, 2012

- 995

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.

- Status
- Not open for further replies.