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?
 A: 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 $$\displaystyle x \sim \pi/4 + \dfrac{0.739085}{2} \sim 1.154940730005\dots$$
See Also: A003957.
A: An analytical form of x can be obtained solving Kepler equation:
$$M= E-\epsilon \sin(E)$$
with eccentricity=1 and mean anomaly = $\pi/2$ by means of Kapteyn series:
$$2x = \frac{\pi}{2}+\sum_{n=1} \frac{2J_n(n)}{n} \sin(\pi n/2)$$
where $J_n()$ are the Bessel functions. Simplifying:
$$2x = \frac{\pi}{2}+\sum_{n=0} \left( \frac{2J_{4n+3}(4n+1)}{4n+1} - \frac{2J_{4n+3}(4n+3)}{4n+3}\right)$$
$$x = \frac{\pi}{4}+\sum_{n=0}  \left( \frac{J_{4n+1}(4n+1)}{4n+1} - \frac{J_{4n+3}(4n+3)}{4n+3}\right)$$ 
Such series can be numerically evaluated, but it converges slowly and n=10000 terms are required to obtain:
$$x = 1.154940317134$$
with 
$$2x-\sin(2x)-\pi/2=-1.38017659479e-006$$
In order to improve the convergence, we can employ an acceleration series technique as Levin's acceleration. (See http://en.wikipedia.org/wiki/Series_acceleration)
With only 10 (ten!) terms we obtain:
$$x=1.1549406884223$$
A simple c++ code, based on gsl library is the following:
#include <iostream>
#include <fstream>
#include <iomanip>
#include "gsl_sf.h"
#include "gsl_sum.h"
using namespace std;

#include <cmath>

int main(int argc, char* argv[])
{
    double PIH = atan(1.)*2;
    cout<<setprecision(13);
    double E=PIH;

    cout<<"raw series"<<endl;
    //raw series
    for( int i = 0 ; i < 1e4; i +=2 )
    {
        double term = 2*gsl_sf_bessel_Jn( 2*i+1, 2*i+1 )/(2*i+1);
        double term2 = 2*gsl_sf_bessel_Jn( 2*i+3, 2*i+3 )/(2*i+3);

        E += (term-term2);
    }
    cout<< E/2<<endl;

    cout<< "error: "<<E-sin(E)-PIH<<endl;

    //levin 
    cout<<"levin accelerated series"<<endl;
    const int N = 10;
    double t[N];
    double sum_accel=0, err;

    gsl_sum_levin_u_workspace* w =
        gsl_sum_levin_u_alloc( N );

    t[0] = PIH;
    for( int i = 1 ; i < N; i++ )
    {
        double term = 2*gsl_sf_bessel_Jn( 4*i-3, 4*i-3 )/(4*i-3);
        double term2 = 2*gsl_sf_bessel_Jn( 4*i-1, 4*i-1 )/(4*i-1);

        t[i] = term-term2;
    }

    gsl_sum_levin_u_accel( t, N, w, &sum_accel, &err );

    E=sum_accel/2;

    cout<<sum_accel/2<<endl;

    cout<<"error: "<<sum_accel-sin(sum_accel)-PIH<<endl;

}

A: 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
$x = \frac{y+\frac{\pi}{2}}{2} = \frac{D}{2} + \frac{\pi}{4}$
A: 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) 
A: As others have noted, $ x = \pi/4 + D/2. $
Based on iteration $ \cos y = y $ there is a fun method to find D using a hand calculator:
Enter any real positive number and press Cos button some 10 or 15 times until none of last digits changes... this is the Dottie number.
A: use $$\sin (2 x)\simeq \sum _{n=0}^{\infty } \frac{i e^{-\frac{i}{2}} (-1)^n x^{n/2} \left((2 i)^n \Gamma \left(n+1,-\frac{i}{2}\right)-e^i (-2 i)^n \Gamma \left(n+1,\frac{i}{2}\right)\right) I_n\left(2 \sqrt{x}\right)}{2 \Gamma (n+1)}$$ few terms gives  good polynomial approximating root
or $$\sin (2 x)\simeq \sum _{n=0}^{\infty } \frac{e^{-\frac{i}{2}} (-1)^n \left(e^i E_{-n}\left(\frac{i}{2}\right)+E_{-n}\left(-\frac{i}{2}\right)\right) x^n \, _0\tilde{F}_1(;n+1;x)}{4 \Gamma (n+1)}$$
