Đáp án:
$1.x=\dfrac{\pi}{9} +\dfrac{k\pi}{3} (k \in \mathbb{Z})$
$2.x = \dfrac{\pi}{4}+\dfrac{k\pi}{2}(k \in \mathbb{Z})$
$3.x \not\in \mathbb{R}(k \in \mathbb{Z})$
$4.x=\dfrac{11\pi}{12}+k\pi(k \in \mathbb{Z})$
Giải thích các bước giải:
1. $\sqrt{3} \tan 3x - 3=0$
ĐKXĐ : $x \ne \dfrac{\pi}{6} + \dfrac{k\pi}{3}$
$\to \tan 3x = \dfrac{3}{\sqrt{3}}$
$\to x=\dfrac{\pi}{9} +\dfrac{k\pi}{3} (k \in \mathbb{Z})$
2. $\tan \Big(3x+\dfrac{\pi}{2} \Big). \cot (5x-\pi) = 1$
ĐKXĐ :$x \ne \dfrac{k\pi}{3}, x \ne \dfrac{k\pi}{5} (k \in \mathbb{Z})$
$\Leftrightarrow - \cot 3x \cot 5x=1$
$\Leftrightarrow \cot 3x \cot 5x+1=0$
$\to \cot(3x - 5x)=0$
$\to x = \dfrac{\pi}{4}+\dfrac{k\pi}{2}$
3. $\cot (4x+2)^2=-\sqrt{3}$
$ x \ne \dfrac{-1}{2} + \dfrac{k\pi}{4} (k \in \mathbb{Z})$
$\to x \not\in \mathbb{R}$
4. $\cot (45°-x)=\dfrac{\sqrt{3}}{3}$
ĐKXĐ $x \ne 45° - k180° (k \in \mathbb{Z})$
$\Leftrightarrow \cot \Big(\dfrac{\pi}{4}-x\Big) = \dfrac{\sqrt{3}}{3}$
$\to \dfrac{\pi}{4}-x=\dfrac{\pi}{3}+k\pi$
$\to x=\dfrac{11\pi}{12}+k\pi$
