2 câu trả lời
Đặt t = tanx
Pt : $2 t^{2}-9 t-11=0$
$\Leftrightarrow\left[\begin{array}{l}t=\frac{11}{2} \\ t=-1\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}\tan x=\frac{11}{2} \\ \tan x=-1\end{array}\right.$
$\left[\begin{array}{l}x=\arctan \frac{11}{2}+k \pi \\ x=\frac{-\pi}{4}+k \pi\end{array} \quad(k \in \mathbb{Z})\right.$
Đáp án:
Đặt `tanx =t`
Ta có phương trình :
$2 t^{2}-9 t-11=0$
$\Leftrightarrow\left[\begin{array}{l}t=\frac{11}{2} \\ t=-1\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}\tan x=\frac{11}{2} \\ \tan x=-1\end{array}\right.$
`⇔`$\left[\begin{array}{l}x=\arctan \frac{11}{2}+k \pi \\ x=\frac{-\pi}{4}+k \pi\end{array} \quad(k \in \mathbb{Z})\right.$
