Eliminate $\theta$ from $\sin3\theta=a\cos\theta$ and $\cos3\theta=b\sin\theta$ 
Eliminate $\theta$ from $$\sin3\theta=a\cos\theta$$ $$\cos3\theta=b\sin\theta$$

I came to this point while trying to solve this problem: Eliminating $\theta$ from trigonometric system (Remark: Symbols differ from the original question)
We can find $$b+a=\frac{2\cos2\theta}{\sin2\theta}$$ $$b-a=\frac{2\cos4\theta}{\sin2\theta}$$
Though at the moment I can't imagine how to arrange these to get a proper relation between $a$ and $b$. We can replace $\sin$ and $\cos$ with $\tan$ and then proceed, but it is cumbersome.
I suspect that there should be a clever way to end this, giving a beautiful answer. For reference, like in this question, a specific method gives the desired answer.$\leftarrow\small \text{(not strictly relevant)}$
So, what is the happy ending of this problem?
 A: My first idea after seeing the problem is to say that
$$
a^2\cos^2\theta+b^2\sin^2\theta=1
$$
From this we get
$$
a^2(1+\cos2\theta)+b^2(1-\cos2\theta)=2
$$
whence
$$
\cos2\theta=\frac{2-a^2-b^2}{a^2-b^2}
$$
But we also have
$$
\frac{\sin^23\theta}{a^2}+\frac{\cos^23\theta}{b^2}=1
$$
whence
$$
b^2\sin^23\theta+a^2\cos^23\theta=a^2b^2
$$
and, similarly to the above,
$$
\cos6\theta=\frac{2a^2b^2-a^2-b^2}{a^2-b^2}
$$
Now use the identity for $\cos3x$ in terms of $\cos x$.
A: I used WolframAlpha to get the following relation between $a$, $b$
$$4 - 3 a^2 - 2 a b + a^3 b - 3 b^2 + 2 a^2 b^2 + a b^3=0$$
which is equivalent to
$$(a-b)^2 = \frac{ ((a+b)^2 -4)^2}{(a+b)^2 + 4}$$
or
$$v^2 = \frac{ (u^2 -1)^2}{u^2 + 1}$$
where $u=\frac{a+b}{2}$ and $v=\frac{a-b}{2}$.
The contour curve is the union of two graphs
$$v = \pm \frac{ u^2 -1}{\sqrt{u^2 + 1}}$$
$\bf{Added:}$ With trigonometry, one  checks that for
$$a = \frac{\sin 3 \theta}{\cos \theta}\\
b= \frac{\cos 3 \theta}{\sin \theta}$$
we have
$$\frac{a+b}{2} = \cot 2 \theta \\
\frac{a b -1}{2} = \cos 4 \theta$$
and from here we get a relation between $a$, $b$
