1. Problem statement: Solve $\sqrt[3]{2\cos^2 x} - \sqrt[3]{2\sin^2 x} = \sqrt[3]{\cos 2x}$ for $x\in[0,\pi]$. 2. Let $a=\sqrt[3]{2\cos^2 x}$, $b=\sqrt[3]{2\sin^2 x}$ and $c=\sqrt[3]{\cos 2x}$. 3. Then $a^3=2\cos^2 x$, $b^3=2\sin^2 x$, $c^3=\cos 2x$. 4. Use the expansions $a^3-b^3=(a-b)(a^2+ab+b^2)$ and $(a-b)^3=a^3-b^3-3ab(a-b)$. 5. Substitute $a-b=c$ and $a^3-b^3=2\cos 2x=2c^3$ into the cubic expansion to obtain $$c^3=2c^3-3abc$$ 6. Rearranging gives $c(c^2-3ab)=0$. 7. Case 1: If $c=0$ then $\cos 2x=0$ and hence $x=\frac{\pi}{4},\frac{3\pi}{4}$ in $[0,\pi]$. 8. Case 2: If $c\neq 0$ then $c^2=3ab$. 9. Compute $a^3b^3=(2\cos^2 x)(2\sin^2 x)=4\cos^2 x\sin^2 x=\sin^2 2x$ so $ab=(\sin^2 2x)^{1/3}=|\sin 2x|^{2/3}$. 10. Thus $c^2=(\cos 2x)^{2/3}=3|\sin 2x|^{2/3}$ and raising both sides to the power $3/2$ gives $|\cos 2x|=3^{3/2}|\sin 2x|$. 11. Use $\sin^2 2x+\cos^2 2x=1$ and set $t=|\cos 2x|$ to solve $t^2+\left(\dfrac{t}{3^{3/2}}\right)^2=1$ which yields $$t^2=\frac{27}{28}\,.$$ 12. Hence $|\cos 2x|=\sqrt{\tfrac{27}{28}}=\frac{3\sqrt3}{2\sqrt7}$ and so $\cos 2x=\pm\frac{3\sqrt3}{2\sqrt7}$. 13. Solve for $x$ on $[0,\pi]$ to get the solutions $\displaystyle x\in\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{1}{2}\arccos\left(\frac{3\sqrt3}{2\sqrt7}\right),\pi-\frac{1}{2}\arccos\left(\frac{3\sqrt3}{2\sqrt7}\right),\frac{1}{2}\arccos\left(-\frac{3\sqrt3}{2\sqrt7}\right),\pi-\frac{1}{2}\arccos\left(-\frac{3\sqrt3}{2\sqrt7}\right)\right\}$. 14. These six values are all solutions in $[0,\pi]$ and we are done.