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