Đáp án:
Giải thích các bước giải:
`sin\ x(sin\ 2x-cos\ x)=0`
`⇔` \(\left[ \begin{array}{l}sin\ x=0\\sin\ 2x-cos\ x=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\2sin\ x . cos\ x-cos\ x=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\cos\ x(2sin\ x -1)=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{2}+k\pi\ (k \in \mathbb{Z})\\sin\ x =\dfrac{1}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{2}+k\pi\ (k \in \mathbb{Z})\\x =\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x =\dfrac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
+) `x = k\pi\ (k \in \mathbb{Z})`
`-\pi \le k\pi \le \pi`
`⇔ -1 \le k \le 1`
Mà `k \in \mathbb{Z}`
`⇒ k \in {-1;0;1}`
+) `x=\frac{\pi}{2}+k\pi\ (k \in \mathbb{Z})`
`-\pi \le \frac{\pi}{2}+k\pi \le \pi`
`⇔ -1 \le \frac{1}{2}+k \le 1`
`⇔ -3/2 \le k \le 1/2`
Mà `k \in \mathbb{Z}`
`⇒ k \in {-1;0}`
+) `x =\frac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})`
`-\pi \le \frac{\pi}{6}+k2\pi \le \pi`
`⇔ -1 \le \frac{1}{6}+2k \le 1`
`⇔ -7/6 \le 2k \le 5/6`
`⇔ -7/12 \le k \le 5/12`
Mà `k \in \mathbb{Z}`
`⇒ k \in {0}`
+) `x =\frac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})`
`-\pi \le \frac{5\pi}{6}+k2\pi \le \pi`
`⇔ -1 \le \frac{5}{6}+2k \le 1`
`⇔ -11/6 \le 2k \le 1/6`
`⇔ -11/12 \le k \le 1/12`
Mà `k \in \mathbb{Z}`
`⇒ k \in {0}`
Vậy có 7 nghiệm thỏa mãn trên `[-\pi;\pi]`
