Mn giúp mk giải pt 4cosx.cos2x+1=0

$4\cos x\cos2x+1=0$ $\Rightarrow 4\cos x(2{\cos}^2x-1)+1=0$ $\Rightarrow 8{\cos}^3x-4\cos x+1=0$ $\Rightarrow \left[ \begin{array}{l} \cos x=\dfrac{1}{2} \\ \cos x=\dfrac{-1-\sqrt5}{4}\\\cos x=\dfrac{-1+\sqrt5}{4} \end{array} \right .$ Với $\cos x=\dfrac{1}{2}\Rightarrow x=\dfrac{\pi}{3}+k2\pi(k\in\mathbb Z)$ Với $\cos x=\dfrac{-1-\sqrt5}{4}\Rightarrow x=\pm\arccos\dfrac{-1-\sqrt5}{4}+k2\pi(k\in\mathbb Z)$ Với $\cos x=\dfrac{-1+\sqrt5}{4}\Rightarrow x=\pm\arccos\dfrac{-1+\sqrt5}{4}+k2\pi(k\in\mathbb Z)$