2) $\sqrt3(\cos 2x+1)+3\sin 2x=3+\sqrt3$
$\Rightarrow \cos 2x+\sqrt3\sin 2x=\sqrt3$
$\Rightarrow \dfrac{1}{2}\cos 2x+\dfrac{\sqrt3}{2}\sin 2x=\dfrac{\sqrt3}{2}$
$\Rightarrow \sin(2x+\dfrac{\pi}{6})=\dfrac{\sqrt3}{2}$
$\Rightarrow \left[ \begin{array}{l} 2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi \\2x+\dfrac{\pi}{6}=\dfrac{2\pi}{3}+k2\pi\end{array} \right .\Rightarrow \left[ \begin{array}{l} x=\dfrac{\pi}{12}+k\pi \\x=\dfrac{\pi}{4}+k\pi\end{array} \right .$
1) Đk: $\sin 2x\ne 0\Leftrightarrow 2x\ne k\pi$
$\Leftrightarrow x\ne k\dfrac{\pi}{2},(k\in\mathbb Z)$
$\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}+4\sin 2x=\dfrac{2}{\sin 2x}$
$\Rightarrow \dfrac{2\cos 2x}{\sin 2x}+4\sin 2x=\dfrac{2}{\sin 2x}$
$\Rightarrow 2\cos 2x+4(1-{\cos}^22x=2)$
$\Rightarrow \left[ \begin{array}{l} \cos 2x=\dfrac{-1}{2}\\ \cos 2x=1 \end{array} \right .\Rightarrow \left[ \begin{array}{l} 2x=\dfrac{\pm2\pi}{3}+k2\pi\\2x=k2pi \end{array} \right .$
$\Rightarrow \left[ \begin{array}{l} x=\dfrac{\pm\pi}{3}+k\pi\\x=k\pi (l)\end{array} \right .(k\in\mathbb Z)$
Vậy $x=\dfrac{\pm\pi}{3}+k\pi,(k\in\mathbb Z)$
