$\begin{array}{l} T8) \,\, 3cos2x - 5cosx - 4 = 0\\ \Leftrightarrow 3(2cos^2x - 1) - 5cosx - 4 = 0\\ \Leftrightarrow 6cos^2x-5cosx-7=0\\ \Leftrightarrow \left[ \begin{array}{l}cosx = \dfrac{5+\sqrt{193}}{12} > 1 \,\,\,(loại)\\cosx = \dfrac{5 - \sqrt{193}}{12} (nhận)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x=arccos\left(\dfrac{5 - \sqrt{193}}{12}\right)+k2\pi\\x=-arccos\left(\dfrac{5 - \sqrt{193}}{12}\right)+k2\pi\end{array} \right. \,\,\, (k \in \Bbb Z)\\\\T10) \,\,cos4x=cos^2x\\ \Leftrightarrow 2cos^22x - 1 = \dfrac{1 + cos2x}{2}\\ \Leftrightarrow 4cos^22x - cos2x - 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l}cos2x=1\\cos2x = -\dfrac{3}{4}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi\\2x =\pm arccos\left(-\dfrac{3}{4}\right)+k2\pi\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi\\x =\pm \dfrac{arccos\left(-\dfrac{3}{4}\right)}{2}+k\pi\end{array} \right.\,\,\,\, (k\in \Bbb Z)\end{array}$
