How to Solve Trigonometric Equations? How are you supposed to go about solving equations such as:
$$-\sqrt{3} = \frac{\sin{4\theta}}{\sin{7\theta}}.$$
I know that $\theta = 30^{\circ}$ is one such solution, but how do I find all solutions using algebra?
Thanks
Edit: I figured out one possible method of reasoning. For $-\sqrt{3} = \frac{\sin{4\theta}}{\sin{7\theta}}$ to be true, taking the "special values" of the $\sin$ function on the unit circle, one way to achieve the value of $-\sqrt{3}$ is to have either $$\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}$$ or $$\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}.$$ Solving for the first case, to achieve a negative value in the denominator, $0\leq\theta\leq\frac{\pi}{4}$ (since you want $4\theta\leq180^{\circ}$ and $\sin{7\theta}\lt0$). Then the only value for $\sin\theta=\frac{\sqrt{3}}{2}$ in the first quadrant is $\theta = \frac{\pi}{6}$. 
Using similar reasoning, you can deduce a symmetrical value in the case where the numerator is negative. This method to me, however, feels unprofessional and "weak." So again, is there a more definitive, algebraic solution?
 A: Use the identities
$$\sin4\theta = 4\cos\theta\left (\sin\theta - 2\sin^3\theta\right)$$
$$ \sin7\theta = -64\sin^7\theta + 112\sin^5\theta -56\sin^3\theta+7\sin\theta,$$ obtained from De Moivre's formula and by taking the substitution $\cos^2\theta
=1-\sin^2\theta$.
Hence,
$$ \frac{4\cos\theta\left (\sin\theta - 2\sin^3\theta\right)}{-64\sin^7\theta + 112\sin^5\theta -56\sin^3\theta+7\sin\theta} = -\sqrt{3} $$$$ \implies \cos\theta = \frac{\sqrt{3}\left(64\sin^7\theta - 112\sin^5\theta +56\sin^3\theta-7\sin\theta\right)}{4\left(\sin\theta - 2\sin^3\theta\right)}.$$
Square the equation (we keep the solutions) and do $y = \sin\theta$. Thus,
$$1-y^2 = \frac{3\left(64y^7-112y^5+56y^3-7y\right)^2}{16\left(y-2y^3\right)^2}$$ $$$$ $$\implies 12288y^{12} -43008y^{10}+59136y^8-40256y^6+13984y^4 -2272y^2 +131 = 0.$$
Now, take $z = y^2$. Hence,
$$12288z^6 -43008z^5+59136z^4-40256z^3+13984z^2 -2272z +131 = 0,$$ which has $6$ positive solutions (found numerically). So, from $z$, you can easily find $y$ and $\theta$.
Remark: Since the equation was squared, you should test the solutions $\theta$ in the original equation. 
