Answer
$${x_1} = 0,{\text{ }}{x_2} \approx - 1.895494,{\text{ }}{x_3} \approx 1.895494$$
Work Step by Step
$$\eqalign{
& y = \sin x,{\text{ }}y = \frac{x}{2} \cr
& {\text{The graphs of the functions are shown below}} \cr
& \cr
& {\text{Find the intersection points}}{\text{, let }}y = y \cr
& \sin x = \frac{x}{2} \cr
& {\text{Subtract }}\frac{x}{2}{\text{ from both sides of equation to write the functions}} \cr
& {\text{in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr
& f\left( x \right) = 0 \cr
& \sin x - \frac{x}{2} = 0 \cr
& {\text{Let }}f\left( x \right) = \sin x - \frac{x}{2},{\text{ differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x - \frac{x}{2}} \right] \cr
& f'\left( x \right) = \cos x - \frac{1}{2} \cr
& {\text{Use the procedure Newton's Method for Approximations}} \cr
& {\text{Roots of }}f\left( x \right) = 0{\text{ }}\left( {{\text{See page 311}}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \cr
& {\text{Substituting }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right){\text{ the Newton's method takes the}} \cr
& {\text{form}} \cr
& {x_{n + 1}} = {x_n} - \frac{{\sin {x_n} - \frac{{{x_n}}}{2}}}{{\cos {x_n} - \frac{1}{2}}} \cr
& {\text{Simplifying}} \cr
& {x_{n + 1}} = {x_n} - \frac{{2\sin {x_n} - {x_n}}}{{2\cos {x_n} - 1}} \cr
& {\text{From the graph we can see that the first intersection point is }} \cr
& \left( {0,0} \right) \cr
& \cr
& {\text{The first possible initial approximation for }}\sin x - \frac{x}{2} = 0{\text{ is}} \cr
& x = - 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& n = 0,{\text{ }}{x_0} = 0 \cr
& {x_{0 + 1}} = {x_0} - \frac{{2\sin {x_0} - {x_0}}}{{2\cos {x_0} - 1}} \cr
& {x_1} \approx - 2 - \frac{{2\sin \left( { - 2} \right) - \left( { - 2} \right)}}{{2\cos \left( { - 2} \right) - 1}} \approx - 1.9000 \cr
& n = 1,{\text{ }}{x_1} = - 1.9000 \cr
& {x_2} \approx - 1.9000 - \frac{{2\sin \left( { - 1.9000} \right) - \left( { - 1.9000} \right)}}{{2\cos \left( { - 1.9000} \right) - 1}} \approx - 1.8955 \cr
& n = 2,{\text{ }}{x_2} = - 1.8955 \cr
& {x_3} \approx - 1.8955 - \frac{{2\sin \left( { - 1.8955} \right) - \left( { - 1.8955} \right)}}{{2\cos \left( { - 1.8955} \right) - 1}} \approx - 1.895494 \cr
& \cr
& {\text{The second possible initial approximation for }}\sin x - \frac{x}{2} = 0{\text{ is}} \cr
& x = 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& n = 0,{\text{ }}{x_0} = 0 \cr
& {x_{0 + 1}} = {x_0} - \frac{{2\sin {x_0} - {x_0}}}{{2\cos {x_0} - 1}} \cr
& {x_1} \approx 2 - \frac{{2\sin \left( 2 \right) - \left( 2 \right)}}{{2\cos \left( 2 \right) - 1}} \approx 1.9000 \cr
& n = 1,{\text{ }}{x_1} = 1.9000 \cr
& {x_2} \approx 1.9000 - \frac{{2\sin \left( {1.9000} \right) - \left( {1.9000} \right)}}{{2\cos \left( {1.9000} \right) - 1}} \approx 1.8955 \cr
& n = 2,{\text{ }}{x_2} = 1.8955 \cr
& {x_3} \approx 1.8955 - \frac{{2\sin \left( {1.8955} \right) - \left( {1.8955} \right)}}{{2\cos \left( {1.8955} \right) - 1}} \approx 1.895494 \cr
& \cr
& {\text{The approximation of the solutions are:}} \cr
& {x_1} = 0,{\text{ }}{x_2} \approx - 1.895494,{\text{ }}{x_3} \approx 1.895494 \cr} $$
