Find polynomial equation for a cardioid in $\mathbb{R}^2$ We have the cardioid with equations:
$$x(\theta)=\cos\theta+\frac{1}{2}\cos(2\theta)$$
$$y(\theta)=\sin\theta+\frac{1}{2}\sin(2\theta)$$
I have to show that you can define this cardioid with a polynomial in two variables.
My approach, I defined the auxiliary variables $u,v$ such that,
$$\begin{matrix}
u=\cos(2\theta)&\Rightarrow&\cos\theta =\frac{\sqrt{1+u}}{2}\\
v=\sin(2\theta)&\Rightarrow&\sin\theta =\frac{\sqrt{1+-u}}{2}
\end{matrix}$$
and since $u^2+v^2=1$, it is,
$$x^2+y^2-\frac{1}{4}=\frac{1}{2}+u\frac{\sqrt{1+u}}{2}+v\frac{\sqrt{1-u}}{2}=ux+vy$$
but I got stuck here, any tips on how can I continue?
 A: Given the two equations
$$x(\theta)=\cos\theta+\frac{1}{2}\cos(2\theta) \tag{1}$$
$$y(\theta)=\sin\theta+\frac{1}{2}\sin(2\theta) \tag{2}$$
By squaring both equations and adding (this gives the useful "expression" $\cos^2+\sin^2$ for both $\theta$ and $2\theta$):
$$\begin{align}
x^2+y^2&=1+\frac{1}{4}+\cos\theta \cos2\theta+\sin\theta\sin2\theta \\[1em]
&=\frac{5}{4}+\cos\theta \tag{3}
\end{align}$$
While from (1) and the identity $\cos2\theta=2\cos^2\theta-1$:
$$\begin{align}
\cos^2\theta+\cos\theta&=x+\frac{1}{2} \\[1em]
\cos^2\theta+\cos\theta+\frac{1}{4}&=x+\frac{3}{4} &(\text{completing the square})\\[1em]
\left(\cos\theta+\frac{1}{2}\right)^2&=x+\frac{3}{4} \tag{4}
\end{align}$$
From (3):
$$\begin{align}
x^2+y^2-\frac{3}{4}&=\cos\theta+\frac{1}{2} &\implies \\[1em]
\left(x^2+y^2-\frac{3}{4}\right)^2&=\left(\cos\theta+\frac{1}{2}\right)^2 &\stackrel{(4)}{\implies} \\[1em]
\left(x^2+y^2-\frac{3}{4}\right)^2&=x+\frac{3}{4}
\end{align}$$
which can be simplified further if you want.
A: Outline: The first equation can be rewritten as 
$$x+\frac{1}{2}=\cos \theta+ cos^2\theta,\tag{1}$$ and the second as 
$$y=\sin\theta(1+\cos\theta).\tag{2}$$ Thus, dividing we find that
$$\tan\theta=\frac{y}{x+\frac{1}{2}}.$$
Square and add $1$. We get $\sec^2\theta$ in terms of $x$ and $y$. So now we have $\cos^2\theta$ in terms of $x$ and $y$.  Rewrite (1) as 
$$cos\theta=\cos^2\theta-x-\frac{1}{2}.$$
Square both sides and substitute for $\cos^2\theta$. Simplify to taste.
