Answer
$$x \approx 4.493409458$$
Work Step by Step
$$\eqalign{
& x - \tan x = 0;{\text{ }}\frac{\pi }{2} < x < \frac{{3\pi }}{2}{\text{ }} \cr
& {\text{From the graph we can see that the equation has infinitely}} \cr
& {\text{solutions}}{\text{, it is a periodic function}}{\text{.}} \cr
& {\text{For the interval }}\frac{\pi }{2} < x < \frac{{3\pi }}{2}{\text{ we have one solution}} \cr
& {\text{Let }}f\left( x \right) = x - \tan x,{\text{ differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {x - \tan x} \right] \cr
& f'\left( x \right) = 1 - {\sec ^2}x \cr
& {\text{Using the Newton's Method }} \cr
& {x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}},{\text{ }}n = 1,2,3,...{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{{x_n} - \tan {x_n}}}{{1 - {{\sec }^2}{x_n}}}{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{From the graph we can see that the possible initial }} \cr
& {\text{approximation is }}{x_1} = 4.5,{\text{ substituting into }}\left( {\bf{2}} \right) \cr
& {x_{n + 1}} = 4.5 - \frac{{\left( {4.5} \right) - \tan \left( {4.5} \right)}}{{1 - {{\sec }^2}\left( {4.5} \right)}} \cr
& {x_2} \approx 4.493613903 \cr
& {\text{Continuing the process we obtain}} \cr
& {x_3} \approx 4.493409655 \cr
& {x_4} \approx 4.493409458 \cr
& {x_5} \approx 4.493409458 \cr
& {x_4} = {x_5}{\text{ into 9 decimal places}}{\text{, then the solution for the given}} \cr
& {\text{interval is }} \cr
& x \approx 4.493409458 \cr} $$
