Solution:

Let $f(x)=x-2 \sin x-3$

$\begin{aligned}\\ & f(0)=-3, f(1)=-2-2 \sin 1, f(2)=-1-2 \sin 2, f(3)=-2 \sin 3, f(4)=1-2 \sin 4 \\\\ & f(-2)=-5+2 \sin 2 \quad, f(-1)=-4+2 \sin 1\\ \end{aligned}\\ $

As $f(3) f(4)\lt0$ by Intermediate value Theorem the root of the real root of the equation $f(x)=0$ lies …