Giải gấp với ạ!!! 4cosx - 2cos2x - cos4x = 1

$4\cos x-2(2{\cos}^2x-1)-(2{\cos}^22x-1)=1$ $\Rightarrow 4\cos x-4{\cos}^2x+2-2{\cos}^22x+1=1$ $\Rightarrow 4\cos x-4{\cos}^2x+2-2(2{\cos}^2x-1)^2=0$ $\Rightarrow 4\cos x-4{\cos}^2x+2-2(4{\cos}^4x-4{\cos}^2x+1)=0$ $\Rightarrow -8{\cos}^4x+4{\cos}^2x+4\cos x=0$ $\Rightarrow 4\cos x(-2{\cos}^2x+\cos x+1)=0$ $\Rightarrow \left[\begin{array}{l} \cos x=0 \\ \cos x=\dfrac{-1}{2} \\ \cos x=1\end{array} \right .\Rightarrow \left[\begin{array}{l}x=\dfrac{\pi}{2}+k\pi \\ x=\dfrac{\pm2\pi}{3}+k2\pi \\ x=k2\pi\end{array} \right .(k\in\mathbb Z)$