Đáp án:
1. \(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{3}+k\pi\end{array} \right.\) 2. \(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{array} \right.\)
3. \(\left[ \begin{array}{l}x=\dfrac{\pi}{8}+k\dfrac{\pi}{2}\\x=\dfrac{arccot 3}{2}+k\dfrac{\pi}{2}\end{array} \right.\) 4.\(\left[ \begin{array}{l}x=arctan2+k\pi\\x=-arctan\dfrac{2}{7}+k\pi\end{array} \right.\)
5. x=±$\dfrac{\pi}{4}+k\pi$ 6. \(\left[ \begin{array}{l}x=\dfrac{7\pi}{12}+k\pi\\x=-\dfrac{\pi}{12}+k\pi\end{array} \right.\)
Giải thích các bước giải:
1, pt ⇔ \(\left[ \begin{array}{l}tan x=-1\\tan x=\sqrt{3}\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{3}+k\pi\end{array} \right.\)
2. pt ⇔ \(\left[ \begin{array}{l}cot x=1\\cot x=-\sqrt{3}\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{array} \right.\)
3. pt ⇔ \(\left[ \begin{array}{l}cot 2x=1\\cot 2x=3\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=\dfrac{\pi}{8}+k\dfrac{\pi}{2}\\x=\dfrac{arc cot 3}{2}+k\dfrac{\pi}{2}\end{array} \right.\)
4. pt ⇔ $7\tan x-\dfrac{4}{\tan x}-12=0$ ( $\tan x\neq0$ ) ⇔ $7\tan^2 x-12\tan x-4=0$
⇔ \(\left[ \begin{array}{l}tan x=-\dfrac{2}{7}\\tan x=2\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=arctan2+k\pi\\x=-arctan\dfrac{2}{7}+k\pi\end{array} \right.\)
5. pt ⇔ $\tan^2 x+\dfrac{1}{\tan^2 x}-2=0$ ⇔ $\tan^4 x-2\tan^2 x+1=0$
⇔ \(\left[ \begin{array}{l}tan x=1\\tan x=-1\end{array} \right.\) ⇔ x=±$\dfrac{\pi}{4}+k\pi$
6. pt ⇔ \(\left[ \begin{array}{l}\tan (2x-\dfrac{\pi}{4})=\sqrt{3}\\\tan (2x-\dfrac{\pi}{4})=-\sqrt{3}\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=\dfrac{7\pi}{12}+k\pi\\x=-\dfrac{\pi}{12}+k\pi\end{array} \right.\)
