Question about converting a polar equation to a rectangular equation $$\sec\theta =2$$
So I went through all the steps and got:
$$\cos\theta =\frac { 1 }{ 2 } $$
$$\sin\theta =\pm \sqrt { 1-\frac { 1 }{ 4 }  } $$
$$\sin\theta =\pm \frac { \sqrt { 3 }  }{ 2 } $$
$$y=\pm {\sqrt { 3 }  }$$
Now why is it that the correct answer has an $x$ in it and why can't I give the answer $x^2+y^2=4$?

From $\sec\theta=2$, draw a triangle in either the first or fourth quadrant with angle $\theta$, adjacent side $1$, opposite side $\sqrt 3$, hypotenuse $2$. Then
\begin{align}
\sec\theta=2\implies \tan\theta&=\pm\sqrt{3}\\
{r\sin\theta\over r\cos\theta}&=\pm \sqrt{3}\\
{y\over x}&=\pm \sqrt{3}\\
y&=\pm \sqrt{3}\,x.
\end{align} 
