Điều kiện: $cos(x)\neq 0\Rightarrow x\neq \frac{\pi}{2}+k\pi(k\in Z)$ptr ban đầu: $2sin(x)cos(x)+\frac{4sin(x)}{cos(x)}=\frac{9\sqrt{3}}{2}$
$\Leftrightarrow 4sin(x)cos^2(x)+8sin(x)=9\sqrt{3}cos(x)$
$\Leftrightarrow 4tan(x)+8tan(x)[1+tan^2(x)]=9\sqrt{3}[1+tan^2(x)](1)$ do ($cos(x)\neq 0$)
Đặt $t=tan(x)$, $(1)$ thành: $8t^3-9\sqrt{3}t^2+12t-9\sqrt{3}=0$
$\Leftrightarrow (t-\sqrt{3})(8t^2-\sqrt{3}t+9)=0$
$\Leftrightarrow t=\sqrt{3}\Rightarrow tan(x)=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi(k\in Z)$
Vậy ptrlg có nghiệm là $x=\frac{\pi}{3}+k\pi(k\in Z)$