Đáp án:
Giải thích các bước giải:
`sin\ x . cos\ x+cos\ x=0`
`⇔ cos\ x(sin\x +1)=0`
`⇔` \(\left[ \begin{array}{l}cos\ x=0\\sin\ x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={\frac{\pi}{2}+k\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})}`
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`(cos\ 2x-cosx).(sinx-1/2)=0`
`⇔` \(\left[ \begin{array}{l} \cos\ 2x-\cos\ x=0\\\sin\ x-\dfrac{1}{2}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l} 2\cos^2 x-1-\cos\ x=0\\\sin\ x=\dfrac{1}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l} (\cos\ x-1)(2\cos\ x+1)=0\\x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l} x=k2\pi\ (k \in \mathbb{Z})\\x=\pm \dfrac{2\pi}{3}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={k2\pi\ (k \in \mathbb{Z});\pm \frac{2\pi}{3}+k2\pi\ (k \in \mathbb{Z});\frac{\pi}{6}+k2\pi\ (k \in \mathbb{Z});\frac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})}`
