Question

1) \(\dfrac{\sqrt{3}\cos x-2}{2\sin x-1}=0\)

2) \(\dfrac{\sqrt{3}\tan x-1}{2\cos x+\sqrt{3}}=0\)

Answer 1

Bài 1:
PT \(\Leftrightarrow \left\{\begin{matrix} \sqrt{3}\cos x=2\\ 2\sin x-1\neq 0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \cos x=\frac{2}{\sqrt{3}}>1\\ \sin x\neq \frac{1}{2}\end{matrix}\right.\) (vô lý do \(\cos x\leq 1\) )

Do đó pt vô nghiệm

Bài 2:

PT \(\Leftrightarrow \left\{\begin{matrix} \tan x=\frac{1}{\sqrt{3}}\\ \cos x\neq \frac{-\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x=\frac{\pi}{6}+k\pi=\frac{-5\pi}{6}+(k+1)\pi\\ x\neq \pm \frac{5}{6}\pi+2t\pi\end{matrix}\right.\)

\(\Rightarrow k+1\) lẻ, hay $k$ chẵn. Do đó:

\( x=\frac{\pi}{6}+2m\pi \) với \(m\in\mathbb{Z}\) nào đó