$\sin \left( {5x} \right) = 2\sin \left( {3x} \right)\sin \left( {4x} \right)$ ask gentlemen to help solve the equation
Where the real number
$$ x \in \mathbb{R}: \sin \left( {5x} \right) = 2\sin \left( {3x} \right)\sin \left( {4x} \right);
 $$
I notice that
$$x = k\pi \quad ;k \in \mathbb{Z}$$
are solutions

Expressing everything in terms of $\sin(x)= s$ and $\cos(x) = c$, the equation says
$$ -64\,{c}^{5}{s}^{2}+16\,{c}^{4}s+48\,{c}^{3}{s}^{2}-12\,{c}^{2}s-8\,c{
s}^{2}+s = 0$$
Taking the resultant of the left side and $c^2 + s^2 - 1$ with respect to $c$, we get 
$$ {s}^{2} \left( 4096\,{s}^{12}-14336\,{s}^{10}+19968\,{s}^{8}-13952\,{s
}^{6}+4976\,{s}^{4}-776\,{s}^{2}+25 \right) = 0
$$
There are $9$ real solutions of this polynomial equation, all of which are in the interval $[-1,1]$.  Except for $0$, I don't think they can be expressed in terms of radicals.  For each nonzero solution $s$, one of the two corresponding values $\pm \sqrt{1-s^2}$ of $\cos(x)$ gives you a solution of the original equation; for $s=0$, both $\cos(x) = \pm 1$ give solutions.  So there are $10$ solutions for $x$ in each interval $[a,a+2\pi)$.
