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