Answer
a. Yes.
b. $-0.979366798,-0.887725837,0.390040317$
Work Step by Step
a. Let $f(x)=x-cos(3x)$; we need to test to see if $f(x)=0$ has at least one solution. We can see that $f(x)$ is a continuous function and $f(0)=-1\lt0$, while $f(\frac{\pi}{3})=\frac{\pi}{3}+1\gt0$. Thus there should be at least one zero in the interval of $(0,\frac{\pi}{3})$. As a matter of fact, there are three solutions as shown in the figure.
b. Use Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with $f'(x)=1+3sin(3x)$ and the starting points $x_0=-1, -0.8,0.5$; we can obtain the three solutions $-0.979366798,-0.887725837,0.390040317$ as shown in the table.
