Đáp án:
Giải thích các bước giải:
`(\sqrt{2}sin\ 2x+2)(2cos\ x+\sqrt{2})=0`
`⇔` \(\left[ \begin{array}{l}\sqrt{2}sin\ 2x+2=0\\2cos\ x+\sqrt{2}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}sin\ 2x=-\dfrac{2}{\sqrt{2}}\\cos\ x=-\dfrac{\sqrt{2}}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}sin\ 2x=-\sqrt{2} < - 1\\cos\ x=cos \dfrac{3\pi}{4}\end{array} \right.\)
`⇔ x=±\frac{3\pi}{4}+k2\pi\ (k \in \mathbb{Z})`
Vậy `S={±\frac{3\pi}{4}+k2\pi\ (k \in \mathbb{Z})}`
b) `(2sin\ x-1)(\sqrt{3}cos\ x-5)=0`
`⇔` \(\left[ \begin{array}{l}2sin\ x-1=0\\\sqrt{3}cos\ x-5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}sin\ x=\dfrac{1}{2}\\cos\ x=\dfrac{5}{\sqrt{3}}>1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\pi-\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={\frac{\pi}{6}+k2\pi\ (k \in \mathbb{Z});\frac{5\pi}{6}+k2\pi\ (k \in \mathbb{Z})}`
