Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{24} + k\dfrac{\pi}{6}\\x = \dfrac{1}{6}\arctan(-6) + k\dfrac{\pi}{6}\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\tan^26x + 5\tan6x - 6 = 0 \qquad (*)$
$ĐKXĐ:\, x \ne \dfrac{\pi}{12} + n\dfrac{\pi}{6}$
$(*)\Leftrightarrow (\tan6x - 1)(\tan6x + 6)=0$
$\Leftrightarrow \left[\begin{array}{l}\tan6x = 1\\\tan6x = -6\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}6x = \dfrac{\pi}{4} + k\pi\\6x = \arctan(-6) + k\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{24} + k\dfrac{\pi}{6}\\x = \dfrac{1}{6}\arctan(-6) + k\dfrac{\pi}{6}\end{array}\right.\quad (k\in\Bbb Z)$
