The difference between the greatest and least values of the function $f(x)=\sin 2 x-x$ on $[-\pi / 2, \pi / 2]$, is

$\frac{\sqrt{3}}{2}+\frac{\pi}{3}$

We have,

$f'(x)=2 \cos 2 x-1$

∴  $f'(x)=0$

$\Rightarrow 2 \cos 2 x-1=0$

$\Rightarrow \cos 2 x=\frac{1}{2} \Rightarrow 2 x=-\pi / 3, \pi / 3 \Rightarrow x=-\pi / 6, \pi / 6$

Now,

$f(-\pi / 2)=\pi / 2, f(\pi / 2)=-\pi / 2$

$f\left(-\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2}+\frac{\pi}{6}$ and $f\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}-\frac{\pi}{6}$

Clearly, $\frac{\sqrt{3}}{2}-\frac{\pi}{6}$ is the greatest value of f(x) and its least value is $-\pi / 2$.

Hence, the required difference is $\frac{\sqrt{3}}{2}-\frac{\pi}{6}-\left(-\frac{\pi}{2}\right)=\frac{\sqrt{3}}{2}+\frac{\pi}{3}$