Answer
$x_5\approx1.16556$
Work Step by Step
Step 1. Given $y=tan(x)$ and $y=2x$, their intersection should satisfy $tan(x)=2x$ with $0\lt x\lt \frac{\pi}{2}$.
Step 2. Letting $f(x)=2x-tan(x)$, we have $f'(x)=2-sec^2x$.
Step 3. Use Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with starting point $x_0=1$ (you can try other starting points in the interval of $(0, \frac{\pi}{2})$ and graphing the function will be great help).
Step 4. Calculating the estimations up to $x_5$ when $f(x_5)$ is very small ($\lt 10^{-5}$ in this case), we can obtain the solution $x_5\approx1.16556$, as shown in the table.
