Đáp án:
$a){\left[\begin{aligned} x=\dfrac{-\pi}{6}+k2\pi\\ x=\dfrac{7\pi}{6}+k2\pi\\ x=\dfrac{\pi}{6}+k\pi \end{aligned}\right.}, (k\in \mathbb{Z})\\
b)
{\left[\begin{aligned} x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\ x=\dfrac{-\pi}{12}+\dfrac{k\pi}{2}\end{aligned}\right.},(k\in \mathbb{Z})\\$
Giải thích các bước giải:
$a) (2\sin x+1)(\cot x-\sqrt{3})=0\\
ĐK: x\neq k\pi,k\in \mathbb{Z}\\
\Leftrightarrow {\left[\begin{aligned}2\sin x+1=0\\\cot x-\sqrt{3}=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2\sin x=-1\\\cot x=\sqrt{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\sin x=\dfrac{-1}{2}\\\dfrac{1}{\tan x}=\sqrt{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{-\pi}{6}+k2\pi\\ x=\pi+\dfrac{\pi}{6}+k2\pi\\ \tan x=\dfrac{1}{\sqrt{3}}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{-\pi}{6}+k2\pi\\ x=\dfrac{7\pi}{6}+k2\pi\\ x=\dfrac{\pi}{6}+k\pi \end{aligned}\right.}, (k\in \mathbb{Z})\\
b)
(\cot2x-1)(3\tan2x+\sqrt{3})=0\\
ĐK: x\neq \dfrac{k\pi}{2};x\neq \dfrac{\pi}{4}+k\pi\\
\Rightarrow {\left[\begin{aligned}\cot2x-1=0\\3\tan2x+\sqrt{3}=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\cot2x=1\\3\tan2x=-\sqrt{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{1}{\tan2x}=1\\\tan2x=\dfrac{-\sqrt{3}}{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\tan2x=1\\\tan2x=\dfrac{-\sqrt{3}}{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 2x=\dfrac{\pi}{4}+k\pi\\ 2x=\dfrac{-\pi}{6}+k\pi\end{aligned}\right.}\\
\Leftrightarrow{\left[\begin{aligned} x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\ x=\dfrac{-\pi}{12}+\dfrac{k\pi}{2}\end{aligned}\right.},(k\in \mathbb{Z})\\$
