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I posted this question all the way back in 2014...and yes, I'm still trying to solve it.

To refresh some memories, here is the little demon in all its glory...

$\displaystyle x \cos \theta + y \sin \theta = \cos (3 \theta)$; $x \sin \theta - y \cos \theta = 3 \sin (3 \theta)$

I actually have some progress...I finally reached some progress when using $$x^2 + y^2 = \cos^2 3 \theta + 9 \sin^2 3 \theta$$ which boils down neatly to $$ 5-(x^2+y^2) = 4 \cos 6 \theta$$ - I then multiplied the first equation by 2 and multiplied that by the second to get

$$(x^2-y^2)(2 \sin \theta \cos \theta) + 2xy (\cos^2 \theta - \sin^2 \theta) = 6 \sin 6 \theta.$$ I would think subsituting $x = \cos \theta$ or $y = \sin \theta$ might give some insight, but I'm not sure.

Is this close? Am I missing something?

(The answer for this is still $\displaystyle(x^2 + y^2)(x^2+y^2+18)+8x(x^2-3y^2) = 27$.)

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    $\begingroup$ How is this not a duplicate of your old question, and what's wrong with the answers you've got there? $\endgroup$ – dxiv Apr 6 '17 at 1:49
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    $\begingroup$ I've updated the original page, linking back to this page. In one of the answers, I got confused by one of the variables, but once I reread the original answer and compared it to this one, that "ah ha" moment came. I will try harder not to "shop" for answers, but this question has been nagging me for three years and I wanted to solve it. I should have gone back to the original rather than generating a new one. $\endgroup$ – bjcolby15 Apr 8 '17 at 21:04
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Here is a solution in outline.

Solve the equations to find $$x = 1 + 2 \cos 2t - 2 \cos^2 2t, y = - 2 \sin 2t ( 1 + \cos 2t).$$

Rewrite the equation for $x$ in terms of $w = 2 \cos 2t - 1$. We find $$3 - 2x = w^2.$$

Now square the equation for $y$ to find $$4y^2 = 16(1 - \cos^2 2t)(1 + \cos 2t)^2 = (1 - w)(3 + w)^3 = -w^4 -8w^3 - 18w^2 + 27,$$ whence $$4y^2 + (3 - 2x)^2 + 18(3 - 2x) - 27 = -8w^3.$$ Now squaring both sides, $$[4y^2 + (3 - 2x)^2 + 18(3 - 2x) - 27]^2 = 64 w^6 = 64 (3 -2x)^3.$$ Dividing by $16$ yields the desired relation.

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  • $\begingroup$ Thank you for your response - this helps a lot. One of the answers is the same as the original post, but your variable opened up the gates up wide. $\endgroup$ – bjcolby15 Apr 8 '17 at 21:07

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