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?