$\sin^3{x}\sin{3x} + \cos^3{x}\cos{3x} = \frac{\sqrt{2}}{4}$
$\Leftrightarrow \dfrac{(3\sin{x} - \sin{3x}) \sin{3x}}{4} + \dfrac{(\cos{3x} + 3\cos{x})\cos{3x}}{4} = \frac{\sqrt{2}}{4}$
$\Leftrightarrow \dfrac{\cos^2{3x} - \sin^2{3x}}{4} + \dfrac{3\cos{2x}}{4} =\frac{\sqrt{2}}{4}$
$\Leftrightarrow \dfrac{\cos{6x} + 3\cos{2x}}{4} =\frac{\sqrt{2}}{4} $
$\Leftrightarrow \cos^3{2x} = \frac{\sqrt{2}}{4}$