4cot^2X +5cotx +1=0 6cos^2x – cosx -1=0 3tan^2x -4tanx +1=0

Đáp án:

$\begin{array}{l}a)\,\,\left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x = \arctan(-4) + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\b)\,\,\left[\begin{array}{l}x = \pi + k2\pi\\x = \pm\arccos\left(-\dfrac{1}{6}\right) + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\c)\,\,\left[\begin{array}{l}x = \dfrac{\pi}{4} + k\pi\\x = \arctan\dfrac{1}{3} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\end{array}$

Giải thích các bước giải:

$\begin{array}{l}a) \,\,4\cot^2x+5\cot x + 1 = 0 \quad (a)\\ ĐKXĐ: \, \sin x \ne 0 \Leftrightarrow x \ne n\pi \quad (n \in \Bbb Z)\\ (a) \Leftrightarrow (\cot x + 1)\left(\cot x + \dfrac{1}{4}\right) =0\\ \Leftrightarrow \left[\begin{array}{l}\cot x = -1\\\cot x = – \dfrac{1}{4}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x = \arctan(-4) + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b)\,\,6\cos^2x – 5\cos x – 1 = 0\\ \Leftrightarrow (\cos x -1)\left(\cos x + \dfrac{1}{6}\right) =0\\ \Leftrightarrow \left[\begin{array}{l}\cos x = 1\\\cos x = – \dfrac{1}{6}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \pi + k2\pi\\x = \pm\arccos\left(-\dfrac{1}{6}\right) + k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ c)\,\,3\tan^2x – 4\tan x + 1 = 0 \quad (c)\\ ĐKXĐ:\, \cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{2} + n\pi \quad (n \in \Bbb Z)\\ (c)\Leftrightarrow (\tan x – 1)\left(\tan x – \dfrac{1}{3}\right) = 0\\ \Leftrightarrow \left[\begin{array}{l}\tan x = 1\\\tan x = \dfrac{1}{3}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k\pi\\x = \arctan\dfrac{1}{3} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ \end{array}$