Answer
$0.450183611$
Work Step by Step
Step 1. Letting $f(x)=2x-cos(x)$, we have $f'(x)=2+sin(x)$.
Step 2. Test signs of $f(x)$, $f(0)=-1\lt0$ and $f(1)=2-cos(1)\gt0$; thus there is a zero in the interval of $(0,1)$
Step 3. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with a starting point $x_0=0$, we can obtain the zero shown in the table as $0.450183611$
