Finding cartesian equation for trigonometric parametric forms I'm trying to find the cartesian equation for these parameteric forms:

$$
x = sin\theta + 2 cos \theta \\
y = 2 sin\theta + cos\theta
$$

I tried:
$$\begin{align}
x^2 & = sin^2\theta + 4cos^2\theta \\
& = 1 - cos^2\theta + 4cos^2\theta \\
& = 1 + 3cos^2\theta \\
\\
y^2 & = 4sin^2\theta+cos^2\theta\\
& = 4(1 - cos^2\theta) + cos^2\theta \\
& = 4 - 3cos^2\theta \\
\\
\therefore & \space4 - y^2 = x^2 - 1\\
\space & x^2 + y^2 = 5
\end{align}$$
Which differs from the given answer of $5x^2 + 5y^2 - 8xy = 9$. Where am I going wrong?

Note that $$x^2\not=\sin^2\theta+4\cos^2\theta$$
and that 
$$x^2=\sin^2\theta+4\cos^2\theta+4\sin\theta\cos\theta$$
because
$$(a+b)^2=a^2+b^2+\color{red}{2ab}.$$
Since we can get
$$\sin\theta=\frac{-x+2y}{3},\ \ \ \cos\theta=\frac{2x-y}{3},$$
then use 
$$\sin^2\theta+\cos^2\theta=1.$$
