Answer
$$ y=-\frac{\sqrt{3}}{2} x+\frac{1}{2}\left(1+\frac{\sqrt{3}\pi}{3}\right).$$
Work Step by Step
Since $ y=\cos x $, then $ y'=-\sin x $ and hence the slope at $ x=\frac{\pi}{3}$ is $ m=-\sin \frac{\pi}{3}=-\frac{\sqrt{3}}{2}.$ Now, the tangent line equation is given by
$$ y=-\frac{\sqrt{3}}{2} x+c.$$
Since the curve and line coincide at $ x=\frac{\pi}{3}$, then
$$\cos \frac{\pi}{3}= -\frac{\sqrt{3}}{2}\frac{\pi}{3}+c \Longrightarrow c=\frac{1}{2}\left(1+\frac{\sqrt{3}\pi}{3}\right)$$
Hence the equation of the tangent line is given by
$$ y=-\frac{\sqrt{3}}{2} x+\frac{1}{2}\left(1+\frac{\sqrt{3}\pi}{3}\right).$$
