Simplify first!

$\begin{aligned} \sin(3x) \sin^3 x + \cos(3x)\cos^3 x & = \sin^6 (2x) \\ \frac {\cos (2x) - \cos (4x)}2 \cdot \frac {1-\cos (2x)}2 + \frac {\cos (2x) + \cos (4x)}2 \cdot \frac {1+\cos (2x)}2 & = \sin^6 (2x) \\ \frac {\cos (2x) + \cos (2x) \cos (4x)}2 & = \sin^6 (2x) \\ \cos (2x) + 2\cos^3 (2x) - \cos (2x) & = 2\sin^6 (2x) \\ \cos^3 (2x) & = \sin^6 (2x) \\ \cos^3 (2x) & = (1 - \cos^2 (2x))^3 \\ \cos (2x) & = 1 - \cos^2 (2x) \\ \cos^2 (2x) + \cos (2x) - 1 & = 0 \\ \cos (2x) & = \frac {\sqrt 5-1}2 \\ 2x & \approx 0.904556894 \\ \implies x & \approx \boxed{0.452} \end{aligned}$

I think there is better solution than mine.

Combine the two equations and using the fact that $\cos{(3x)}=6\cos^3{(u)}-3\cos{(u)}$ , we get

The second factor have no real solution. Using quadratic formula and because $|y|\leq1$ , we get the solution from the first factor