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!
 A: 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 $(A)$
A: Here's a brute force approach to #2. For simplicity, I'll write $s$ and $c$ for $\sin\theta$ and $\cos\theta$.


*

*Expand $\cos 3\theta$ and $\sin 3\theta$ $$\begin{align} 
x c + y s &= c \; ( 4 c^2 - 3 ) \\
x s - y c &= s \; ( 4 c^2 - 1 )
\end{align}$$

*Isolate $s$ terms, and square, so that we can re-write in terms of $c$. $$\begin{align}
(y s)^2 &= c^2 ( 4 c^2 - 3 - x )^2 \quad \to \quad y^2 ( 1 - c^2 ) - c^2 ( 4 c^2 - 3 - x )^2 = 0\\
(y c)^2 &= s^2 ( 4 c^2 - 1 - x )^2 \quad \to \quad y^2 c^2 - ( 1 - c^2 )( 4 c^2 - 1 - x )^2 = 0
\end{align}$$

*Invoke one of my favorite tools ---the "method of resultants" (which I describe a bit in this answer)--- to eliminate $c$. Mathematica's (and/or WolframAlpha's) Resultant[] function makes it easy, yielding this polynomial equation: $$\begin{align}(\;x^4 - 4 y^4 + 3 x^2 y^2 + 8 x^3 - 18 x y^2 + 18 x^2  + 27 y^2 - 27 \;) & \\ \cdot\;(\;x^4 + y^4 + 2 x^2 y^2 + 8 x^3 - 24 x y^2 + 18 x^2 + 18 y^2 - 27 \;) &= 0\end{align}$$

*The second factor corresponds to the answer you expect. Presumably, the first factor is extraneous in the current context (as is often the case with resultant results), but it's not obvious to me why that is.
The same process works for #1.
