Đáp án: \(\left[ \begin{array}{l}x=\dfrac{7\pi}{27}+\dfrac{k2\pi}{3}\\x=-\dfrac{5\pi}{27}+\dfrac{k2\pi}{3}\end{array} \right.\) $(k∈\mathbb{Z})$
Giải thích các bước giải:
$cos(3x-\dfrac{\pi}{9})=-\dfrac{1}{2}$
⇔ $cos(3x-\dfrac{\pi}{9})=cos\dfrac{2\pi}{3}$
⇔ \(\left[ \begin{array}{l}3x-\dfrac{\pi}{9}=\dfrac{2\pi}{3}+k2\pi\\3x-\dfrac{\pi}{9}=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}3x=\dfrac{7\pi}{9}+k2\pi\\3x=-\dfrac{5\pi}{9}+k2\pi\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\dfrac{7\pi}{27}+\dfrac{k2\pi}{3}\\x=-\dfrac{5\pi}{27}+\dfrac{k2\pi}{3}\end{array} \right.\) $(k∈\mathbb{Z})$
