Đáp án:
Giải thích các bước giải:
`cos\ (2x-\frac{\pi}{6})=-1/2`
`⇔ cos\ (2x-\frac{\pi}{6})=cos\ (\frac{2\pi}{3})`
`⇔` \(\left[ \begin{array}{l}2x-\dfrac{\pi}{6}=\dfrac{2\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\) `(k \in \mathbb{Z})`
`⇔` \(\left[ \begin{array}{l}2x=\dfrac{5\pi}{6}+k2\pi\\2x=\dfrac{-\pi}{2}+k2\pi\end{array} \right.\) `(k \in \mathbb{Z})`
`⇔` \(\left[ \begin{array}{l}x=\dfrac{5\pi}{12}+k\pi\\x=\dfrac{-\pi}{4}+k\pi\end{array} \right.\) `(k \in \mathbb{Z})`
+) `x=\frac{5\pi}{12}+k\pi\ (k \in \mathbb{Z})`
`\frac{-\pi}{2} < \frac{5\pi}{12}+k\pi < \frac{8\pi}{3}`
`⇔ \frac{-1}{2} < \frac{5}{12}+k < \frac{8}{3}`
`⇔ \frac{-11}{12} < k < \frac{9}{4}`
`k \in \mathbb{Z} ⇒ k \in {0;1;2;3;4}`
`⇒ x \in {\frac{5\pi}{12};\frac{17\pi}{12};\frac{29\pi}{12};\frac{41\pi}{12};\frac{53\pi}{12}}`
+) `x=\frac{-\pi}{4}+k\pi\ (k \in \mathbb{Z})`
`\frac{-\pi}{2} <\frac{-\pi}{4}+k\pi< \frac{8\pi}{3}`
`⇔ \frac{-1}{2} < \frac{-1}{4}+k < \frac{8}{3}`
`⇔ \frac{-1}{4} < k < \frac{35}{12}`
`k \in \mathbb{Z} ⇒ k \in {0;1;2}`
`⇒ x \in {\frac{-\pi}{4};\frac{3\pi}{4};\frac{7\pi}{4}}`
