Find the value of $x$ in the figure

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 $x = 6^\circ$.

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.