<s>[INST] Find the value of $x$ in the figure 
Attempt: $\frac{\sin5x}{\sin8x} = \frac{\sin2x}{\sin3x}$
$x= 6^\circ$ 
(The answer is $6^\circ$)
How to ensure there is no other solution?
 [/INST] 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 $x = 6^\circ$.
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