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Eliminate $\theta$ from the equation

$$x\cos(3\theta)+y\sin(3\theta)=a\cos(\theta)\\ x\sin(3\theta)+y\cos(3\theta)=a\cos(\theta+\frac{\pi}{6})$$

I tried squaring a adding but got nowhere. Also got $x$ and $y$ as linear equations in form of $\theta$ but cant see what to do next.

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Not elegant at all!

Squaring & adding $$\dfrac{x^2+y^2+2xy\sin6\theta}{a^2}= \cos^2\theta+\cos^2\left(\theta+\dfrac\pi6\right) =1+\cos\dfrac\pi6\cos\left(2\theta+\dfrac\pi6\right)\ \ \ \ (1)$$

using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

Adding & squaring $$\dfrac{(x+y)^2(1+\sin6\theta)}{a^2}= \left[\cos\theta+\cos\left(\theta+\dfrac\pi6\right)\right]^2=\left(1+\cos\dfrac\pi6\right)\left(1+\cos\left(2\theta+\dfrac\pi6\right)\right)\ \ \ \ (2)$$

using Prosthaphaeresis Formula $\cos C+\cos D$ and Double angle formula

Set $2\theta+\dfrac\pi6=t,6\theta=3t-\dfrac\pi2$

From $(1),$ $$\dfrac{x^2+y^2-2xy\cos3t}{a^2}=1+\cos\dfrac\pi6\cos t\ \ \ \ (3)$$

From $(2),$ $$\dfrac{(x+y)^2(1-\cos3t)}{a^2}=\left(1+\cos\dfrac\pi6\right)\left(1+\cos t\right)\ \ \ \ (4)$$

Solve $(3),(4)$ for $\cos t,\cos3t$

and use $\cos3t=4\cos^3t-3\cos t$ to eliminate $t$

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