## Calculus (3rd Edition)

$$y=-\frac{\sqrt{3}}{2} x+\frac{1}{2}\left(1+\frac{\sqrt{3}\pi}{3}\right).$$
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 $$y=-\frac{\sqrt{3}}{2} x+\frac{1}{2}\left(1+\frac{\sqrt{3}\pi}{3}\right).$$