How to Solve Trigonometric Equations?

Question

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?

Answer 1

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.

Answer 2

You can write the equation as $$\frac{\sin 4\theta}{\sin \theta} + \sqrt{3}\frac{\sin 7\theta}{\sin \theta} = 0.$$ Each fraction can be written as a function of $t = \cos \theta$. This shows that whether an angle is a solution or not depends only on its cosine, hence you need only find the solutions $\theta$ to the equation on $(0,\pi)$, and then any other solutions will be obtained as $\pm \theta + 2\pi n$.

Expicitly, we find $$8t^3 - 4t + \sqrt{3}(64t^6 - 80t^4 + 24t^2 - 1) = 0.$$ Since $t_1 = \sqrt{3}/2$ is a solution, we divide by $2t-\sqrt{3}$ and find the equation $$32 \sqrt{3} t^5+48 t^4-16 \sqrt{3} t^3-20 t^2+2 \sqrt{3} t+1 = 0,$$ which according to Wolfram Alpha has the solutions $$ t_2 \approx -0.93531434645348586217 \\ t_3 \approx -0.64480000754256621365 \\ t_4 \approx -0.17020647383131240361 \\ t_5 \approx 0.30168315890275447056 \\ t_6 \approx 0.58261226514017136211 \\ $$ To find all the solutions to the original equation in $(0,\pi)$, we must take the arccosines of these numbers, in addition to $\arccos t_1 = \pi/6$. Thus the solutions are $\pm \theta_i + 2\pi n$, where $$ \theta_1 = \pi/6 \\ \theta_2 \approx 2.7799427920394627406 \\ \theta_3 \approx 2.2715579312242782072 \\ \theta_4 \approx 1.7418355233567214888 \\ \theta_5 \approx 1.2643387521053641769 \\ \theta_6 \approx 0.94885721910009571098 $$

Unfortunately, I have no idea whether these numbers have closed forms.