Đặt $2x = a$, $\dfrac{8x^{3}+2001}{2002}=b$.
Khi đó ta có hệ: $\left\{\begin{matrix} b^{3}=2002a-2001\\a^{3}=2002b-2001 \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{array}{l}(a-b)(a^{2}+ab+b^{2}+2002)=0\\a^{3}=2002b-2001\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}a=b\\a^{3}=2002b-2001\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}a=b=1\\ a=b=\dfrac{-1+\sqrt{8005}}{2}\\ a=b=\dfrac{-1-\sqrt{8005}}{2}\end{array}\right.$
Từ đó suy ra: $x\in\{\dfrac{1}{2};\dfrac{-1+\sqrt{8005}}{4};\dfrac{-1-\sqrt{8005}}{4}\}$.