<s>[INST] Two trigonometric eliminations I have a couple of trigonometric elimination questions for you...
$1) a \sin \theta = b \sin (2 \theta)$; $c \cos \theta = d \cos (2 \theta)$
*NOTE: the first equation was incorrect...I had $\cos 2 \theta$ and it should be $\sin 2 \theta$.
I was able to get the answer to $2(abc - a^2 d + 2b^2 d)=0$, but the actual answer is $(c^2 - d^2)(abc - a^2 d + 2b^2 d)=0$.  Where does the $(c^2 -d^2)$ come from?
UPDATE: I think I know $(c^2-d^2)$ comes from.  If $\theta$ =0, then cos $\theta$ = 1 and in turn cos 2 $\theta$ = 1; plugging these values into the first equation gives us $c = d$. (I already did the second condition where $\frac {a} {2b}$))  If we square the first condition and subtract, that gives us $(c^2-d^2)$; since this also equals $2(abc - a^2 d + 2b^2 d)=0$, we can multiply both sides by $(c^2-d^2)$ to get the final result.  
UPDATE #2: Actually, the more I thought about it, this is correct.
$2) \displaystyle x \cos \theta + y \sin \theta = \cos (3 \theta)$; $x \sin \theta - y \cos \theta = 3 \sin (3 \theta)$
I'm not sure where to start...I've tried squaring both equations, multiplying, etc.  (The answer for this is $\displaystyle(x^2 + y^2)(x^2+y^2+18)+8x(x^2-3y^2) = 27$.)
UPDATE: I was able to get the following for x and y (thanks to the suggestion!)
$\displaystyle x= 2\cos 2 \theta - \cos 4 \theta$
$\displaystyle y = 2\sin 2 \theta + \sin 4 \theta$
UPDATE 2/16/14: BREAKTHROUGH!  Never mind...that answer bombed out.
UPDATE 2/17/14: Not giving up and using the advice below, I finally got $\displaystyle c = \frac {(x-1)^2 + (y^2-4)}{12}$...I'm going to plug this in to the equation for x and see how it goes.  If all works well, I should finally get the answer, save all the cleanup work.  Never mind.
UPDATE 4/8/17: I reposted question #2 to get a fresh perspective...see Trigonometric elimination (reprise from 2014). If it's a duplicate, my apologies!
Thanks for your help!
 [/INST] HINT:
$1)$ For the original version:
Divide to find $\tan\theta$ 
and  use $\displaystyle\cos2A=\frac{1-\tan^2A}{1+\tan^2A}$ 
either  $(i)$ in square one of the given equation
or $(ii)$ in $$\left(\frac{b\cos2\theta}b\right)^2+\left(\frac{d\cos2\theta}c\right)^2=\sin^2\theta+\cos^2\theta=\cdots$$
For the edited version
From the first relation, $\displaystyle\sin\theta(a-2b\cos\theta)=0$
If $\displaystyle\sin\theta=0,\cos\theta=\pm1,\cos2\theta=2\cos^2\theta-1=1$
Else $\displaystyle\cos\theta=\frac a{2b}$
Put the values of $\displaystyle\cos\theta,\cos2\theta$ in the second relation, $\displaystyle c\cos\theta=d\cos2\theta$

$2)$ 
Solving for $x,y$
$\displaystyle x=2\cos2\theta-\cos4\theta=2c-(2c^2-1)\iff 2c^2=2c-1-x\  \ \ \ (A)$ where $c=\cos2\theta$
$\displaystyle y=2\sin2\theta-\sin4\theta=2\sin2\theta(1-\cos2\theta)$
$\displaystyle\implies y^2=4(1-c^2)(1-c)^2\ \ \ \ (B)$
Put the value of $2c^2$ from $(A)$ in $(B)$ to find $c$ in terms of $y^2$
Then, put this value of $c$ in $(A)$
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