The equation
$$\sin(3x)\sin(4x)=\cos(4x)\cos(5x)\tag{1}$$
can be transformed by expanding in powers of $\cos(x)$ and letting $r=\cos(x)$;
$$128r^9-256r^7+160r^5-32r^3+r=0,$$
‘factor out’ the solution $r=0$;
$$128r^8-256r^6+160r^4-32r^2+1=0,$$
letting $s=2r$;
$$8s^4-32s^3+40s^2-16s+1=0,$$
letting $t=s-1$;
$$8t^4-8t^2+1=0,$$
letting $u=t^2$;
$$8u^2-8u+1=0,$$
which can be solved using the quadratic formula.
Going back through the steps gives the solutions to (1) as
$$x=\pm\arccos(...)+k\cdot2\pi$$
(18 solutions + period of $2\pi$).

Is there a simpler way to solve this equation?