Đáp án:
Giải thích các bước giải:
4) `y=\sqrt{\frac{sin\ x+2}{cos\ x+1}}`
ĐK: `\frac{sin\ x+2}{cos\ x+1} \ge 0`
Ta có: `-1 \le sin\ x \le 1`
`⇔ 1 \le sin\ x+2 \le 3`
`⇒ sin\ x+2 \ge 0 ∀x`
`⇔ cos\ x \ne -1`
`⇔ x \ne \pi+k2\pi\ (k \in \mathbb{Z})`
Vậy `D=\mathbb{R} \\ {\pi+k2\pi\ (k \in \mathbb{Z})}`
6) `y=\frac{2}{cos\ x-cos\ 3x}`
ĐK: `cos\ x \ne cos\ 3x`
`⇔` \(\left[ \begin{array}{l}x \ne 3x+k2\pi\ (k \in \mathbb{Z})\\x \ne -3x+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x \ne -k\pi\ (k \in \mathbb{Z})\\x \ne k\dfrac{\pi}{2}\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `D=\mathbb{R} \\ {-k\pi\ (k \in \mathbb{Z});k\frac{\pi}{2}\ (k \in \mathbb{Z})}`
