Answer
$r\displaystyle \cos(\theta+\frac{\pi}{6})=\frac{1}{2}$
Work Step by Step
Conversion formulas:
$\left\{\begin{array}{ll}
(x,y)=(r\cos\theta,r\sin\theta) & \\
r^{2}=x^{2}+y^{2}, & \tan\theta=\frac{y}{x}
\end{array}\right.$
Replace x and y with $ r\cos\theta$ and $ r\sin\theta$
$\sqrt{3}r\cos\theta-r\sin\theta=1$
$r(\sqrt{3}\cos\theta-\sin\theta)=1$
Divide with 2, because $\displaystyle \frac{\sqrt{3}}{2}$ and $\displaystyle \frac{1}{2}$ are possible cosine and sine values.
$r(\displaystyle \frac{\sqrt{3}}{2}\cos\theta-\frac{1}{2}\sin\theta)=\frac{1}{2}$
We recognize the angle of $\displaystyle \frac{\pi}{6}$
$r(\displaystyle \cos\frac{\pi}{6}\cos\theta-\sin\frac{\pi}{6}\sin\theta)=\frac{1}{2}$
We recognize the parentheses from the additive identity for cosine,
$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
The polar equation is
$r\displaystyle \cos(\theta+\frac{\pi}{6})=\frac{1}{2}$
