# Solving $2x - \sin 2x = \pi/2$ for $0 < x < \pi/2$

What is $x$ in closed form if $2x-\sin2x=\pi/2$, $x$ in the first quadrant?

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Yes. But it seems from other posters' answers, there is no known one due to $cosy=y$. –  Robert Nov 15 '10 at 19:56

The solution is given by $$\displaystyle x = \pi/4 + D/2$$

where $\displaystyle D$ is the root of $\cos y = y$

The root of $\displaystyle \cos y = y$ is nowadays known as the Dottie Number and apparently has no known "closed form" solution. If you consider this number to be part of your constants, then the above can be considered a closed form solution.

For a proof:

If $\displaystyle y = \sin(2x)$

then we have that

$\displaystyle 2x = \pi/2 + y$

$\displaystyle y = \sin 2x = \sin (\pi/2 + y) = \cos y$.

The root of $$\displaystyle y = \cos y$$ is $\displaystyle y = 0.739085\dots$

Notice that $\displaystyle \pi/2 > x \gt \pi/4$ (as $\displaystyle f(x) = 2x - \sin 2x$ is increasing in $\displaystyle [0,\pi/2]$), so if $\displaystyle x = \pi/4 + z$ then

$\displaystyle \sin(2x) = \sin(\pi/2 + 2z) = \cos 2z = 0.739085\dots$

And thus $\displaystyle z = \dfrac{0.739085\dots}{2}$.

Thus $$\displaystyle x \sim \pi/4 + \dfrac{0.739085}{2} \sim 1.154940730005\dots$$

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+1 for explaining what the Dottie Number is. –  Américo Tavares Nov 16 '10 at 0:49

Repeating the same thing as the other answers, but with a moderately more elegant one:

Let $y = 2x - \frac{\pi}{2}$, then substituting:

$y = sin(y+\frac{\pi}{2})$

$y = cos(y)$

$y = D$

then

$x = \frac{y+\frac{\pi}{2}}{2} = \frac{D}{2} + \frac{\pi}{4}$

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To get a graphical impression of the solution have a look here:

http://www.wolframalpha.com/input/?i=solve+2+x+-+sin+(2+x)+%3D+Pi/2+for+x

Edit: That WA doesn't give a closed form is at least a hint that there might be none (very probable)

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If you assume the Dottie number to be part of the "closed form" constants, then there is a closed form solution. See my answer. –  Aryabhata Nov 15 '10 at 19:35
@Moron: Well, I don't know if there is some "degree of closeness" but I guess this way you could define everything as "closed form" (which is often done btw, just think of Pi, e, Si(x), Li(x) asf... –  vonjd Nov 15 '10 at 19:42
Not sure what you mean. All I am saying is that it can be written in terms of well known constants. I guess we are just agreeing :-) –  Aryabhata Nov 15 '10 at 19:45
@Moron & @vonjd: You're both correct. As I said somewhere here previously, "closed-form" is just shorthand for "popular enough to be given a name and notation." –  Ｊ. Ｍ. Nov 15 '10 at 23:16