# 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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