Đáp án:
$\left[ \begin{array}{l}x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{3\pi}{4}+k\pi\end{array} \right. $($k∈\mathbb{Z}$)
Giải thích các bước giải:
Tập xác định: $D=\mathbb{R}$
$4sin\Bigg(2x-\dfrac{\pi}{6}\Bigg)+2\sqrt[]{3}=0$
$↔ 4sin\Bigg(2x-\dfrac{\pi}{6}\Bigg)=-2\sqrt[]{3}$
$↔ sin\Bigg(2x-\dfrac{\pi}{6}\Bigg)=-\dfrac{\sqrt[]{3}}{2}$
$↔ sin\Bigg(2x-\dfrac{\pi}{6}\Bigg)=sin\Bigg(-\dfrac{\pi}{3}\Bigg)$
$↔ \left[ \begin{array}{l}2x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=\dfrac{4\pi}{3}+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{3\pi}{4}+k\pi\end{array} \right. $($k∈\mathbb{Z}$)
