#### Answer

The equation of the tangent line at $\theta=\frac{\pi}{3}$ is given by
$$ y=(1-\sqrt{3}) \theta+\sqrt{3}+1-\frac{\pi(1-\sqrt{3})}{3} .$$

#### Work Step by Step

Since $ y=2(\sin \theta+\cos \theta)$, then one can find that $ y'=2(\cos \theta-\sin \theta)$ and hence the slope at $\theta=\frac{\pi}{3}$ is $ m=2(\cos \frac{\pi}{3}-\sin \frac{\pi}{3})=1-\sqrt{3}.$ Now, the equation of the tangent line at $\theta=\frac{\pi}{3}$ is given by
$$ y=(1-\sqrt{3}) \theta+c.$$
Since the curve and the line coincide at $\theta=\frac{\pi}{3}$, then
$$2(\sin \frac{\pi}{3}+\cos \frac{\pi}{3})=(1-\sqrt{3}) \frac{\pi}{3}+c \Longrightarrow c= \sqrt{3}+1-\frac{\pi(1-\sqrt{3})}{3} .$$
Hence the equation of the tangent line at $\theta=\frac{\pi}{3}$ is given by
$$ y=(1-\sqrt{3}) \theta+\sqrt{3}+1-\frac{\pi(1-\sqrt{3})}{3} .$$