Đáp án:
$\left[\begin{array}{l} x=100^\circ+ k 180^\circ(k \in \mathbb{Z})\\ x=-40^\circ+ k 90^\circ(k \in \mathbb{Z})\end{array} \right.$
Giải thích các bước giải:
$\cos(3x-20^\circ)=-\cos x\\ \Leftrightarrow \cos(3x-20^\circ)=\cos( x+180^\circ)\\ \Leftrightarrow \left[\begin{array}{l} 3x-20^\circ= x+180^\circ+ k 360^\circ(k \in \mathbb{Z})\\ 3x-20^\circ=- x-180^\circ+ k 360^\circ(k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} 2x=200^\circ+ k 360^\circ(k \in \mathbb{Z})\\ 4x=-160^\circ+ k 360^\circ(k \in \mathbb{Z})\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=100^\circ+ k 180^\circ(k \in \mathbb{Z})\\ x=-40^\circ+ k 90^\circ(k \in \mathbb{Z})\end{array} \right.$
