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 between 3 and 4

Let $\mathrm{x}_0=4$ be the initial guess to the equation (2).

Then $x_1=x_0-\left[f\left(x_0\right) / f^{\prime}\left(x_0\right)\right]=2-f(2) / f^{\prime}(2)=3.09900$

$ \begin{aligned}\\ & x_2=x_1-\left[f\left(x_1\right) / f^{\prime}\left(x_1\right)\right]=-1.099-f(-1.099) / f^{\prime}(-1.099)=3.10448 \\\\ & x_3=x_2-\left[f\left(x_2\right) / f^{\prime}\left(x_2\right)\right]=3.10450 \\\\ & x_4=x_3-\left[f\left(x_3\right) / f^{\prime}\left(x_3\right)\right]=3.10451\\ \end{aligned}\\ $