Đáp án:
Giải thích các bước giải:
`sin2x + cosx = 0`
`⇔ 2sin\ x . cos\ x+cos\ x=0`
`⇔ cos\ x(2sin\ x+1)=0`
`⇔` \(\left[ \begin{array}{l}cos\ x=0\\2sin\ x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
+) `x=\frac{\pi}{2}+k\pi\ (k \in \mathbb{Z})`
`0<\frac{\pi}{2}+k\pi<2\pi`
`⇔ 0<\frac{1}{2}+k<2`
`⇔ -1/2 < k < 3/2`
Mà `k \in \mathbb{Z}`
`⇒ k \in {0,1}`
+) `x=\frac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})`
`0<\frac{\pi}{6}+k2\pi<2\pi`
`⇔ 0<\frac{1}{6}+2k<2`
`⇔ -1/6 < 2k < 11/6`
`⇔ -1/12 < k < 11/12`
Mà `k \in \mathbb{Z}`
`⇒ k \in {0}`
+) `x=\frac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})`
`0<\frac{5\pi}{6}+k2\pi<2\pi`
`⇔ 0<\frac{5}{6}+2k<2`
`⇔ -5/6 < 2k < 7/6`
`⇔ -5/12 < k < 7/12`
Mà `k \in \mathbb{Z}`
`⇒ k \in {0}`
Vậy .........
