Trigonometry by Jose Sacramento

$\begin{aligned} \arccos 2x - \arccos x & = \frac \pi 3 \\ \cos (\arccos 2x - \arccos x) & = \cos \frac \pi 3 \\ 2x^2 + \sqrt{(1-4x^2)(1-x^2)} & = \frac 12 \\ 2\sqrt{(1-4x^2)(1-x^2)} & = 1 - 4x^2 \\ 4(1-4x^2)(1-x^2) & = (1 - 4x^2)^2 \\ 4(1-4x^2)(1-x^2 - 1 + 4x^2) & = 0 \\ x^2(1-4x^2) & = 0 \end{aligned}$

$\implies x = \begin{cases} 0 & \implies \arccos 0 - \arccos 0 = 0 \ne \dfrac \pi 3 & {\color{#D61F06}\text{rejected}} \\ \dfrac 12 & \implies \arccos 1 - \arccos \dfrac 12 = - \dfrac \pi 3 \ne \dfrac \pi 3 & {\color{#D61F06}\text{rejected}} \\ -\dfrac 12 & \implies \arccos (-1) - \arccos \left( -\dfrac 12 \right) = \dfrac \pi 3 & {\color{#3D99F6}\text{accepted}} \end{cases}$

$\implies x = \boxed{-0.5}$