Giải giúp mình 4 câu này với ạ. Mình cần gấp lắm ạ.

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:

$2\tan^2x – 5\tan – 1 = 0$

$\Leftrightarrow \left[\begin{array}{l}\tan x = \dfrac{5 +\sqrt{33}}{4}\\\tan x = \dfrac{5 -\sqrt{33}}{4}\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = \arctan\left(\dfrac{5 +\sqrt{33}}{4}\right) + k\pi\\ x = \arctan\left(\dfrac{5 -\sqrt{33}}{4}\right) + k\pi\end{array}\right.\quad (k\in\Bbb Z)$