Elimination of Trigonometric Functions Is there a simple way to eliminate the trigonometric functions here? 
$$
\begin{array}{lcl}
A\cos(3\omega\tau)+B\sin(3\omega\tau)+C\cos(\omega\tau) &=& D\\
B\cos(3\omega\tau)+A\sin(3\omega\tau)+C\sin(\omega\tau) &=& 0
\end{array}
$$
I would ideally like to solve for $\tau$. Please help!
 A: The standard way to eliminate trig functions is the Weierstrass substitution. Let $t = \tan\tfrac12\theta$. Then
$$
\cos\theta = \frac{1-t^2}{1+t^2} \quad ; \quad \sin\theta = \frac{2t}{1+t^2}
$$
In your case, you will also need the facts:
$$
\sin 3\theta = 3\sin\theta -4\sin^3\theta    \\
\cos 3\theta = -3\cos\theta +4\cos^3\theta
$$
This will eliminate the trig functions, but it looks like the resulting equations will still be difficult to solve.
A: Let only $C\cos(\omega\tau)$ in the lhs of the first equation, $C\sin(\omega\tau)$in the lhs of the second equation,square lhs's and rhs's and add the two equations. You should arrive to 
$$C^2=A^2+B^2+D^2-2 D (A \cos (3 \omega\tau)+B \sin (3 \omega\tau))+4 A B \sin (3 \omega\tau) \cos (3 \omega\tau)$$ At least you now only have a "simpler" formula since now only angle $3 \omega\tau$ appears in it.  
If now, you use the Weierstrass substitution as suggested by bubba, you end with a polynomial of $t = \tan \left(\frac{3 \tau  \omega }{2}\right)$ (probably a fourth order polynomial).
