Answer
$-0.739085133$
Work Step by Step
Step 1. Letting $f(x)=x+cos(x)$, we have $f'(x)=1-sin(x)$. Graphing the function, we know there is only one zero.
Step 2. Test signs of $f(x)$, $f(0)=1\gt0$ and $f(-\frac{\pi}{2})=-\frac{\pi}{2}+0\lt0$; thus there is a zero in the interval of $(-\frac{\pi}{2},0)$
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 $-0.739085133$
