Find the value of $x$ in the figure

Question

Attempt: $\frac{\sin5x}{\sin8x} = \frac{\sin2x}{\sin3x}$

$x= 6^\circ$

(The answer is $6^\circ$)

How to ensure there is no other solution?

Answer 1

Let $t = e^{ix}$. We then have $\sin nx = \frac{1}{2i}(t^n - t^{-n})$.

Expanding everything in the identity $\frac{\sin5x}{\sin8x} = \frac{\sin2x}{\sin3x}$, we have an equation: $$t^{20} - t^{18} - t^{16} + t^{12} + t^8 - t^4 - t^2 + 1 = 0,$$ which after factorization becomes: $$(t - 1)^2 (t + 1)^2 (t^{16} + t^{14} - t^{10} - t^8 - t^6 + t^2 + 1) = 0.$$

Of course, $t = \pm 1$ are not solutions to our problem. Thus $t$ is a root of the polynomial $t^{16} + t^{14} - t^{10} - t^8 - t^6 + t^2 + 1$, which happens to be exactly the $60$-th cyclotomic polynomial $\Phi_{60}(t)$.

Therefore $t$ is one of the $60$-th primitive roots of unity, and hence $x$ is equal to $6k^\circ$ for some $k$ prime to $60$.

But from the graph, we should have $5x < 180^\circ$, hence $k < 6$. This only leaves the possibility $k = 1$, or $x = 6^\circ$.

Answer 2

This isn't a complete answer, but I would like to know how I could complete it.

I thought of a more "linear" approach, pun intended, than WhatsUp's responce : Using the Law of sines we can write (from $ADB$) $\ \frac {sin(2x)}{DB}= \frac {sin(180-3x)}{AB}$ and also (from $BDC$) $\ \frac {sin(5x)}{DB}= \frac {sin(180-8x)}{DC}$. \begin{cases} \frac {sin(2x)}{DB} &= \,\frac {sin(180-3x)}{AB} \\ \frac {sin(5x)}{DB} &= \frac {sin(180-8x)}{DC}\\ \end{cases} Dividing both equations and using the fact that $DC=AB\ $ we get:

$\frac {sin(2x)}{sin (5x)} = \frac {sin(180-3x)}{sin(180-8x)}\Leftrightarrow \frac {sin(2x)}{sin (5x)} = \frac {sin(3x)}{sin(8x)} \Leftrightarrow sin(2x)sin(8x)=sin(3x)sin(5x) \Leftrightarrow \\ \frac{1}{2} ({cos(2x − 8x) − cos(2x + 8x)})= \frac{1}{2} ({cos(3x − 5x) − cos(3x + 5x)}) \Leftrightarrow \\ cos(-6x)-cos(10x) = cos(-2x)-cos(8x) \Leftrightarrow cos(-6x)+cos(8x)=cos(-2x)+cos(10x) $

If we try x=6 the equality holds, but I now I have got the same problem as Tortugut, are there anymore possibilities? Is there a way to continue this approach, (using trigonometric iddentities until we solve for x), or we must use Euler's identity and factor it like WhatsUp did?

Answer 3

Note that, with $t=\cos2x$, the equation

$$\frac{\sin5x}{\sin8x} = \frac{\sin2x}{\sin3x}$$

is to be factorized as,

$$(t-1)\left(t-\cos\frac\pi{15}\right)\left(t-\cos\frac{7\pi}{15}\right)\left(t-\cos\frac{11\pi}{15}\right)\left(t-\cos\frac{13\pi}{15}\right)=0$$

which yields its explicit solutions. But, only $x=\frac\pi{30}$ is valid for the problem given.