Đáp án:
Giải thích các bước giải:
7) `cos\ (2x+\frac{\pi}{4})+2=0`
`⇔ cos\ (2x+\frac{\pi}{4})=-2`
Ta có: `cos\ x \in [-1;1]` mà `-2 < -1` nên PT vô nghiệm
8) `cos\ (\frac{x}{3}-30^{0})=1`
`⇔ \frac{x}{3}-30^{0}=k360^{0}\ (k \in \mathbb{Z})`
`⇔ \frac{x}{3}=30^{0}+k360^{0}\ (k \in \mathbb{Z})`
`⇔ x=90^{0}+k1080{0}\ (k \in \mathbb{Z})`
Vậy `S={90^{0}+k1080{0}\ (k \in \mathbb{Z})}`
9) `sin^2 2x=1/4`
`⇔` \(\left[ \begin{array}{l}\sin\ 2x=\dfrac{1}{2}\\\sin\ 2x=-\dfrac{1}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}\sin\ 2x=\sin\ \dfrac{\pi}{6}\\\sin\ 2x=\sin\ -\dfrac{\pi}{6}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\2x=\pi-\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\2x=-\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\2x=\pi+\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{12}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{12}+k\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{12}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{7\pi}{12}+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy .......
