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