Answer
$x-\sqrt{3}y=-6$
.
Work Step by Step
Conversion formulas:
$\left\{\begin{array}{ll}
(x,y)=(r\cos\theta,r\sin\theta) & \\
r^{2}=x^{2}+y^{2}, & \tan\theta=\frac{y}{x}
\end{array}\right.$
Use the additive identity for cosine
$\displaystyle \cos(\theta-\frac{2\pi}{3})=\cos\theta\cos\frac{2\pi}{3}+\sin\theta\sin\frac{2\pi}{3}$
$\displaystyle \cos(\theta-\frac{2\pi}{3})=-\frac{1}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta\qquad/\times r$
$ r\displaystyle \cos(\theta-\frac{2\pi}{3})=-\frac{1}{2}r\cos\theta+\frac{\sqrt{3}}{2}r\sin\theta$
Apply the conversion formula
$\quad (x,y)=(r\cos\theta,r\sin\theta)$
$r\displaystyle \cos(\theta-\frac{\pi}{4})=-\frac{1}{2}x+\frac{\sqrt{3}}{2}y$
In Cartesian coordinates, the line equation is
$-\displaystyle \frac{1}{2}x+\frac{\sqrt{3}}{2}y=3\qquad/\times(-2)$
$x-\sqrt{3}y=-6$
