Đáp án:
$a)x=\pm\dfrac{5\pi}{12} + k \pi( k \in \mathbb{Z})\\ b)x=75^\circ+ k 180^\circ( k \in \mathbb{Z}).$
Giải thích các bước giải:
$a)2\cos 2x+\sqrt{3}=0\\ \Leftrightarrow 2\cos 2x=-\sqrt{3}\\ \Leftrightarrow \cos 2x=-\dfrac{\sqrt{3}}{2}\\ \Leftrightarrow \cos 2x=\cos \dfrac{5\pi}{6}\\ \Leftrightarrow 2x=\pm\dfrac{5\pi}{6} + k 2 \pi( k \in \mathbb{Z})\\ \Leftrightarrow x=\pm\dfrac{5\pi}{12} + k \pi( k \in \mathbb{Z})\\ b)\tan(x-15^\circ)-\sqrt{3}=0\\ \Leftrightarrow \tan(x-15^\circ)=\sqrt{3}\\ \Leftrightarrow \tan(x-15^\circ)=\tan 60^\circ\\ \Leftrightarrow x-15^\circ=60^\circ+ k 180^\circ( k \in \mathbb{Z})\\ \Leftrightarrow x=75^\circ+ k 180^\circ( k \in \mathbb{Z}).$
