1) $3\tan^2x - 2\sqrt\tan x + \sqrt3= 0\qquad (1)$
$ĐKXĐ:\, x \ne \dfrac{\pi}{2} + n\pi\quad (n\in \Bbb Z)$
$(1)\Leftrightarrow \left[\begin{array}{l}\tan x = 1 \\tan x = \dfrac{\sqrt3}{3}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}x = \dfrac{\pi}{4} + k\pi\\x = \dfrac{\pi}{6}+k\pi\end{array}\right.\quad (k\in\Bbb Z)$
2) $\sqrt3\tan x + 6\cot x + 2\sqrt3 - 3 = 0\qquad (2)$
$ĐKXĐ:\, x \ne n\dfrac{\pi}{4}\quad (n\in \Bbb Z)$
$(2) \Leftrightarrow \sqrt3\tan^2x + (2\sqrt3 - 3)\tan x -6 = 0$
$\Leftrightarrow \left[\begin{array}{l}\tan x = -2\\\tan x = \sqrt3\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \arctan(-2)+k\pi\\x = \dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
3) $3\cos^26x + 8\sin3x\cos3x - 4 = 0$
$\Leftrightarrow 3(1 - \sin^26x) + 4\sin6x - 4 = 0$
$\Leftrightarrow 3\sin^26x - 4\sin6x + 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin6x = 1\\\sin6x = \dfrac{1}{3}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}6x = \dfrac{\pi}{2} + k2\pi\\6x = \arcsin\dfrac{1}{3} + k2\pi\\6x =\pi -\arcsin\dfrac{1}{3} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k\dfrac{\pi}{3}\\x = \dfrac{1}{6}\arcsin\dfrac{1}{3} + k\dfrac{\pi}{3}\\x =\dfrac{\pi}{6} -\dfrac{1}{6}\arcsin\dfrac{1}{3} + k\dfrac{\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$
4) $2\sin^2x - 5\sin x\cos x - \cos^2x =0$
Nhận thấy $\cos x = 0$ không là nghiệm của phương trình
Chia 2 vế của phương trình cho $\cos^2x$ ta được:
$2\tan^2x - 5\tan - 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}\tan x = \dfrac{5 +\sqrt{33}}{4}\\\tan x = \dfrac{5 -\sqrt{33}}{4}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \arctan\left(\dfrac{5 +\sqrt{33}}{4}\right) + k\pi\\ x = \arctan\left(\dfrac{5 -\sqrt{33}}{4}\right) + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
