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?

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