The greatest value of  $y=\sin 2x-x$ on $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ is

$y^{\text{'}}=2\cos 2x-1\Rightarrow y^{\text{'}}\left(x\right)=0\iff {\cos }^{2}x=\frac{1}{2}$

$\text{ So, }{y}^{\text{'}}\left(x\right)=0\text{ if }x=\frac{\pi }{6}\text{. }$ Thus the critical points of $y$ in

$\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\text{ are }-\frac{\pi }{2},\frac{\pi }{6},\frac{\pi }{2}\text{. Now }y\left(-\frac{\pi }{2}\right)=\frac{\pi }{2},y\left(\frac{\pi }{2}\right)=-\frac{\pi }{2}$

and $y\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}-\frac{\pi }{2}$ The greatest value is $\frac{\pi }{2}$.