Đáp án:
$1, tan²2x - 5tan2x + 6 = 0 $
$⇔$ \(\left[ \begin{array}{l}tan2x=3\\tan2x=2\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\frac{1}{2}arc tan3+\frac{k\pi}{2}\\x=\frac{1}{2}arctan2+\frac{k\pi}{2}\end{array} \right.\)
$2, 2cos²3x - cos3x - 1 = 0$
$⇔$ \(\left[ \begin{array}{l}cos3x=1\\cos3x=\frac{-1}{2}\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\frac{k2\pi}{3}\\x=±\frac{2\pi}{9}+\frac{k2\pi}{3}\end{array} \right.\)
