Question 12.7: A root of 3 sin x = x is near to x = 2.5. Use two iterations......

The equation must first be written in the form f (x) = 0, that is

f(x)=3 \sin x-x=0

Then

\begin{array}{ll}x_1=2.5 & \\f(x)=3 \sin x-x & f\left(x_1\right)=-0.705 \\f^{\prime}(x)=3 \cos x-1 & f^{\prime}\left(x_1\right)=-3.403\end{array}

Then

x_2=2.5-\frac{(-0.705)}{(-3.403)}=2.293

The process is repeated with  x_1  = 2.293 as the initial approximation:

x_1=2.293 \quad f\left(x_1\right)=-0.042 \quad f^{\prime}\left(x_1\right)=-2.983

Then

x_2=2.293-\frac{(-0.042)}{(-2.983)}=2.279

Using two iterations of the Newton–Raphson technique, we obtain x = 2.28 as an improved estimate of the root.