$f_1\left(x\right)=2x,f_2\left(x\right)=3\sin x-x\cos x$ then for $x\in (0,\frac{\pi }{2})$

Let$h\left(x\right)=f_1\left(x\right)-f_2\left(x\right)$

$=2x-(3\sin x-x\cos x)$

$=2x-3\sin x+x\cos x$

$h\text{'}\left(x\right)=2-3\cos x-x\sin x+\cos x$

$=2-2\cos x-x\sin x$

$h"\left(x\right)=2\sin x-x\cos x-\sin x=\sin x-x\cos x$

$h"\text{'}\left(x\right)=\cos x+x\sin x-\cos x=x\sin x$

$h"\text{'}\left(x\right)>0\Rightarrow h"\left(x\right)>0\Rightarrow h\text{'}\left(x\right)>0\Rightarrow h\left(x\right)>0$