Answer
\[\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{1.500000} \\
1&{1.443890} \\
2&{1.361976} \\
3&{1.268175} \\
4&{1.196179} \\
5&{1.168571} \\
6&{1.165592} \\
7&{1.165561} \\
8&{1.165561} \\
9&{1.165561} \\
{10}&{1.165561}
\end{array}}\]
Work Step by Step
\[\begin{gathered}
{\text{Let }}f\left( x \right) = \tan x - 2x,{\text{ and }}{x_0} = 1.5 \hfill \\
{\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}} \hfill \\
f'\left( x \right) = {\sec ^2}x - 2 \hfill \\
{\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \hfill \\
{x_{n + 1}} = {x_n} - \frac{{\tan {x_n} - 2{x_n}}}{{{{\sec }^2}{x_n} - 2}} \hfill \\
\hfill \\
{\text{Taking }}{x_0} = 1.5 \hfill \\
{x_0} = 1.5 \hfill \\
{x_{0 + 1}} = {x_1} = 1.5 - \frac{{\tan \left( {1.5} \right) - 2\left( {1.5} \right)}}{{{{\sec }^2}\left( {1.5} \right) - 2}} \approx 1.443890 \hfill \\
{x_{1 + 1}} = {x_2} = 1.443890 - \frac{{\tan \left( {1.443890} \right) - 2\left( {1.443890} \right)}}{{{{\sec }^2}\left( {1.443890} \right) - 2}} \approx 1.361976 \hfill \\
{x_{1 + 2}} = {x_3} = 1.361976 - \frac{{\tan \left( {1.361976} \right) - 2\left( {1.361976} \right)}}{{{{\sec }^2}\left( {1.361976} \right) - 2}} \approx 1.268175 \hfill \\
{x_{1 + 3}} = {x_4} = 1.268175 - \frac{{\tan \left( {1.268175} \right) - 2\left( {1.268175} \right)}}{{{{\sec }^2}\left( {1.268175} \right) - 2}} \approx 1.196179 \hfill \\
{x_{1 + 4}} = {x_5} = 1.196179 - \frac{{\tan \left( {1.196179} \right) - 2\left( {1.196179} \right)}}{{{{\sec }^2}\left( {1.196179} \right) - 2}} \approx 1.168571 \hfill \\
{x_{1 + 5}} = {x_6} = 1.168571 - \frac{{\tan \left( {1.168571} \right) - 2\left( {1.168571} \right)}}{{{{\sec }^2}\left( {1.168571} \right) - 2}} \approx 1.165592 \hfill \\
{x_{1 + 6}} = {x_7} = 1.165592 - \frac{{\tan \left( {1.165592} \right) - 2\left( {1.165592} \right)}}{{{{\sec }^2}\left( {1.165592} \right) - 2}} \approx 1.165561 \hfill \\
{x_{1 + 7}} = {x_8} = 1.165561 - \frac{{\tan \left( {1.165561} \right) - 2\left( {1.165561} \right)}}{{{{\sec }^2}\left( {1.165561} \right) - 2}} \approx 1.165561 \hfill \\
{x_{1 + 8}} = {x_9} = 1.165561 - \frac{{\tan \left( {1.165561} \right) - 2\left( {1.165561} \right)}}{{{{\sec }^2}\left( {1.165561} \right) - 2}} \approx 1.165561 \hfill \\
{x_{1 + 9}} = {x_{10}} = 1.165561 - \frac{{\tan \left( {1.165561} \right) - 2\left( {1.165561} \right)}}{{{{\sec }^2}\left( {1.165561} \right) - 2}} \approx 1.165561 \hfill \\
\hfill \\
{\text{Thus}}{\text{, we obtain}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
n&{{x_n}} \\
0&{1.500000} \\
1&{1.443890} \\
2&{1.361976} \\
3&{1.268175} \\
4&{1.196179} \\
5&{1.168571} \\
6&{1.165592} \\
7&{1.165561} \\
8&{1.165561} \\
9&{1.165561} \\
{10}&{1.165561}
\end{array}} \hfill \\
\end{gathered} \]
