Đáp án:
$x=\dfrac{5\pi}{24}+\dfrac{k \pi}{2}(k \in \mathbb{Z}).$
Giải thích các bước giải:
$\tan \left(3x-\dfrac{\pi}{4}\right)=\tan \left(x+\dfrac{\pi}{6}\right)\\ \Leftrightarrow 3x-\dfrac{\pi}{4}=x+\dfrac{\pi}{6}+k \pi(k \in \mathbb{Z})\\ \Leftrightarrow 2x=\dfrac{5\pi}{12}+k \pi(k \in \mathbb{Z})\\ \Leftrightarrow x=\dfrac{5\pi}{24}+\dfrac{k \pi}{2}(k \in \mathbb{Z}).$