Đáp án:
Giải thích các bước giải:
`sin^2 x+1/2 sin\ 2x+sin\ x=0`
`⇔ sin^2 x+1/2 . 2 sin\ x . cos\ x+sin\ x=0`
`⇔ sin^2 x+sin\ x . cos\ x+sin\ x=0`
`⇔ sin\ x(sin\ x+cos\ x+1)=0`
`⇔` \(\left[ \begin{array}{l}\sin\ x=0\\\sin\ x+cos\ x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}\sin\ x=0\\\sin\ x+cos\ x=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\\sqrt{2}\sin\ (x+\dfrac{\pi}{4})=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\\sin\ (x+\dfrac{\pi}{4})=-\dfrac{1}{\sqrt{2}}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\x+\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=k\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\\x=\pi+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
