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 $x^2+y^2=4$?

$$ \sec \theta = 2 $$
$$ 1 = 2 \cos \theta $$
$$ r = 2 r\cos \theta $$
$$ \sqrt{x^2 + y^2} = 2x $$
$$ x^2 + y^2 = 4x^2 $$
$$ 3x^2 = y^2 $$
$$ y = \pm \sqrt{3} \,x$$
Another method: 
$$ \cos \theta = \frac{1}{2} \Rightarrow \tan \theta = \pm \sqrt{3} = \frac{y}{x} \Rightarrow y = \pm \sqrt{3} \,x $$
