$3\cos 4x - 2\cos^23x = 1\Leftrightarrow 3cos4x-(cos6x+1)=1$
$cos6x-3cos4x+2=0$
Đặt $t=cos2x(-1\leq t\leq 1)$
Ta được : $(4t^3-3t)-3(2t^2-1)+2=0$
$\Leftrightarrow 4t^3-6t^2-3t+5=0\Leftrightarrow (t-1)(4t^2-2t-5)=0$
$\Leftrightarrow \left[\begin{array}{I}
t=1\\ t= \frac{1\pm\sqrt{21}}{4}\end{array}\right.$
$\left[\begin{array}{I}
\cos 2x=1\\ \cos 2x=\frac{1\pm\sqrt{21}}{4}(\cos 2x=\frac{1+\sqrt{21}}{4}(L))\end{array}\right.$
$\Leftrightarrow \left[ \begin{array}{l}x =k\pi\\x=\pm\frac{1}{2}\alpha +k\pi \end{array} \right. $ Với $\alpha$ là góc mà $cos\alpha=\frac{1-\sqrt{21} }{4} $