Answer
$$x \approx 1.029866529$$
Work Step by Step
$$\eqalign{
& 2\cos x = x;{\text{ }}x > 0{\text{ }} \cr
& {\text{From the graph we can see that the equation has one solution }} \cr
& {\text{Subtract }}x{\text{ from both sides of equation to write the functions}} \cr
& {\text{in the form }}f\left( x \right) = g\left( x \right) - h\left( x \right) = 0 \cr
& 2\cos x - x = 0 \cr
& {\text{Let }}f\left( x \right) = 2\cos x - x,{\text{ differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {2\cos x - x} \right] \cr
& f'\left( x \right) = - 2\sin x - 1 \cr
& {\text{Using the Newton's Method }} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}},{\text{ }}n = 1,2,3,...{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{2\cos {x_n} - {x_n}}}{{ - 2\sin {x_n} - 1}}{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{From the graph we can see that the possible initial }} \cr
& {\text{approximation is }}{x_1} = 1,{\text{ substituting into }}\left( {\bf{2}} \right) \cr
& {x_{n + 1}} = \left( 1 \right) - \frac{{2\cos \left( 1 \right) - \left( 1 \right)}}{{ - 2\sin \left( 1 \right) - 1}} \cr
& {x_2} \approx 1.030043368 \cr
& {\text{Continuing the process we obtain}} \cr
& {x_3} \approx 1.029866535 \cr
& {x_4} \approx 1.029866529 \cr
& {x_5} \approx 1.029866529 \cr
& {x_4} = {x_5}{\text{ into 9 decimal places}}{\text{, then the solution for }}x > 0{\text{ of}} \cr
& {\text{the equation is }} \cr
& x \approx 1.029866529 \cr} $$
