The interval on which the function sin2x–x, $0<x<\frac{\pi }{2}$
decreases is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\sin 2\mathrm{x}-\mathrm{x} \\ {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=\left(\cos 2\mathrm{x}\right)\left(2\right)-1 \end{gathered}$$

$\mathrm{f}\left(\mathrm{x}\right)$ is decreases

$$\begin{gathered} \Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)<0 \\ 2\cos 2\mathrm{x}-1<0 \\ \cos 2\mathrm{x}<1/2 \\ 2\mathrm{x}>\frac{\mathrm{\pi }}{3} \\ \mathrm{x}>\frac{\mathrm{\pi }}{6} \\ \therefore \mathrm{x}\in \left(\frac{\mathrm{\pi }}{6}, \frac{\mathrm{\pi }}{2}\right) \\ \mathrm{Becasue} \mathrm{given} \mathrm{x}\in \left(0, \frac{\mathrm{\pi }}{2}\right) \end{gathered}$$