MHB Is the equation $2x-1-\sin x=0$ solvable with exactly one real solution?

[Chris L T521]

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:

Spoiler

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]

This week's problem was correctly answered by anemone, eddybob123, johng, and MarkFL. You can find anemone's solution below.

[sp]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 $$\frac{5 \pi}{18}$$ and $$\frac{\pi}{30}$$. 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 $$\frac{5 \pi}{18}$$ and $$\frac{\pi}{30}$$ 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.[/sp]