Đáp án:
Giải thích các bước giải:
`tan 3x - 4cot 3x +3=0`
`⇔ tan 3x -4.\frac{1}{tan 3x}+3=0`
`⇔ tan^2 3x-4+3tan 3x=0`
`⇔ (tan 3x-1)(tan 3x-4)=0`
`⇔` \(\left[ \begin{array}{l}tan\ 3x=1\\tan\ 3x=4\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{12}+k\dfrac{\pi}{3}\ (k \in \mathbb{Z})\\x=\dfrac{arctan (4)}{3}+k\dfrac{\pi}{3}\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={\frac{\pi}{12}+k\frac{\pi}{3}\ (k \in \mathbb{Z});\frac{arctan (4)}{3}+k\frac{\pi}{3}\ (k \in \mathbb{Z})}`
