Đáp án:
Giải thích các bước giải:
`sin\ 2x=-1/2`
`⇔ sin\ 2x=sin\ (-\frac{\pi}{6})`
`⇔` \(\left[ \begin{array}{l}2x=-\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\2x=\pi+\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{\pi}{12}+k\pi\ (k \in \mathbb{Z})\\2x=\dfrac{7\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{\pi}{12}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{7\pi}{12}+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`0 < x < \pi`
`⇔ 0 < -\frac{\pi}{12}+k\pi < \pi`
`⇔ 0 < -\frac{1}{12}+k < 1`
`⇔ 1/12 < k < 13/12`
Mà `k \in \mathbb{Z} ⇒ k = 1`
`⇒ x=\frac{11\pi}{12}`
`⇔ 0 < \frac{7\pi}{12}+k\pi < \pi`
`⇔ 0 < \frac{7}{12}+k < 1`
`⇔ -7/12 < k < 5/12`
Mà `k \in \mathbb{Z} ⇒ k = 0`
`⇒ x=-\frac{\pi}{12}`
Vậy có 2 nghiệm TM
