The normal form of line $\sqrt{3}x+y+4=0$ is

$\sqrt{3}x+y=-4$      $divide with \sqrt{a^2+b^2}=\sqrt{3+1} =2$

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

                   $$\begin{gathered} \cos \alpha =-\frac{\sqrt{3}}{2} and \sin \alpha = \frac{-1}{2} \\ \alpha \in Third quadrant \Rightarrow \alpha =\mathrm{\pi } +\frac{\mathrm{\pi }}{6} \end{gathered}$$

                $\Rightarrow x\cos \frac{7\pi }{6}+y\sin \frac{7\pi }{6}=2$