giải pt: 16/ `4cos^2x +3tan^2x-4\sqrt{3}cosx+2\sqrt{3}tanx+4=0`

Đáp án:

 $x = -\dfrac{\pi}{6} + k2\pi \quad (k\in\Bbb Z)$

Giải thích các bước giải:

$4\cos^2x +3\tan^2x-4\sqrt{3}\cos x+2\sqrt{3}\tan x+4=0$ $(*)$

$ĐKXĐ: \, x \ne \dfrac{\pi}{2} + n\pi\quad (n\in \Bbb Z)$

$(*) \Leftrightarrow \left(4\cos^2x – 2.4\cos x.\sqrt3 + 3\right) + \left(3\tan^2x + 2.\sqrt3\tan x+ 1\right) = 0$

$\Leftrightarrow (2\cos x – \sqrt3)^2 + (\sqrt3\tan x + 1)^2 = 0$

$\Leftrightarrow \begin{cases}2\cos x – \sqrt3 = 0\\\sqrt3\tan x + 1 = 0\end{cases}$

$\Leftrightarrow \begin{cases}\cos x = \dfrac{\sqrt3}{2}\\\tan x = -\dfrac{1}{\sqrt3}\end{cases}$

$\Leftrightarrow \begin{cases}x = \pm \dfrac{\pi}{6} + k2\pi\\x = -\dfrac{\pi}{6} + k\pi\end{cases}$

$\Leftrightarrow x = -\dfrac{\pi}{6} + k2\pi \quad (k\in\Bbb Z)$