Đáp án:
Giải thích các bước giải:
`sin\ (2x+pi/3)=0`
`⇔ 2x+pi/3=k\pi\ (k \in \mathbb{Z})`
`⇔ 2x=-pi/3+k\pi\ (k \in \mathbb{Z})`
`⇔ x=-pi/6+k\frac{\pi}{2}\ (k \in \mathbb{Z})`
Vậy `S={-pi/6+k\frac{\pi}{2}\ (k \in \mathbb{Z})}`
`cos\ (3x-\pi/4)=1`
`⇔ 3x-\pi/4=k2\pi\ (k \in \mathbb{Z})`
`⇔ 3x=\pi/4+k2\pi\ (k \in \mathbb{Z})`
`⇔ x=\pi/12+k\frac{2\pi}{3}\ (k \in \mathbb{Z})`
Vậy `S={\pi/12+k\frac{2\pi}{3}\ (k \in \mathbb{Z})}`
`sin\ x=sin\ \frac{\pi}{12}`
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{12}+k2\pi\ (k \in \mathbb{Z})\\x=\pi-\dfrac{\pi}{12}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{12}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{11\pi}{12}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={\frac{\pi}{12}+k2\pi\ (k \in \mathbb{Z});\frac{11\pi}{12}+k2\pi\ (k \in \mathbb{Z})}`
`cos\ x=2/3`
`⇔ x=\pm arccos\ (2/3)+k2\pi\ (k \in \mathbb{Z})`
Vậy `S={\pm arccos\ (2/3)+k2\pi\ (k \in \mathbb{Z})}`
