A trig for the day

$\sin(X^\circ) =4\sin(20^\circ) \sin(40^\circ) \sin(80^\circ) \\ = 2\sin(80^\circ) \cdot 2\sin(20^\circ) \sin(40^\circ) \\ =2\sin(80^\circ) \cdot (\cos(20^\circ) - cos(60^\circ)) \\ = 2\sin(80^\circ) \cdot (\cos(20^\circ) - \dfrac{1}{2}) \\ =2\sin(80^\circ)\cos(20^\circ) - \sin(80^\circ) \\ = (\sin(100^\circ) + \sin(60^\circ)) - \sin(80^\circ) \\ = \sin(\pi - 80) + \sin(60^\circ)) - \sin(80^\circ) = \sin(80^\circ) +\sin(60^\circ) - \sin(80^\circ) = \sin(60^\circ) \\ \therefore X^\circ = \large\boxed{60^\circ}$

Note that

$4\sin(x)\sin(60^{\circ} - x)\sin(60^{\circ} + x) = \sin(x)(\sqrt{3}\cos(x) - \sin(x))(\sqrt{3}\cos(x) + \sin(x)) =$

$\sin(x)(3\cos^{2}(x) - \sin^{2}(x)) = \sin(x)(3 - 4\sin^{2}(x)) = 3\sin(x) - 4\sin^{3}(x) = \sin(3x)$ ,

since

$\sin(3x) = \sin(2x + x) = \sin(2x)\cos(x) + \cos(2x)\sin(x) =$

$2\sin(x)\cos^{2}(x) + (1 - 2\sin^{2}(x))\sin(x) = 2\sin(x)(1 - \sin^{2}(x)) + \sin(x) - 2\sin^{3}(x) =$

$3\sin(x) - 4\sin^{3}(x)$ .

So with $x = 20^{\circ}$ we have that $4\sin(20^{\circ})\sin(40^{\circ})\sin(80^{\circ}) = \sin(3 \times 20^{\circ}) = \sin(60^{\circ})$ ,

and thus $X = \boxed{60}$ .