Answer
The graphs of the curve and its tangent lines are below.
Work Step by Step
$$y=f(x)=1+\cos x\hspace{1cm}\{-3\pi/2\le x\le 2\pi\}$$
The derivative of $f(x)$: $$f'(x)=0-\sin x=-\sin x$$
The tangent lines to $f(x)$ are:
1) At $x=-\pi/3$:
$f(x)=1+\cos(-\pi/3)=1+\frac{1}{2}=\frac{3}{2}$
$f'(x)=-\sin(-\pi/3)=-(-\frac{\sqrt3}{2})=\frac{\sqrt3}{2}$
$$y-\frac{3}{2}=\frac{\sqrt3}{2}(x+\frac{\pi}{3})=\frac{\sqrt3}{2}x+\frac{\sqrt3\pi}{6}$$ $$y=\frac{\sqrt3}{2}x+\frac{\sqrt3\pi}{6}+\frac{3}{2}=\frac{\sqrt3}{2}x+\frac{\sqrt3\pi+9}{6}$$
2) At $x=3\pi/2$:
$f(x)=1+\cos(3\pi/2)=1+0=1$
$f'(x)=-\sin(3\pi/2)=-(-1)=1$
$$y-1=1(x-\frac{3\pi}{2})=x-\frac{3\pi}{2}$$ $$y=x-\frac{3\pi}{2}+1=x+\frac{2-3\pi}{2}$$
The graphs of the curve and its tangent lines are below.
