Đáp án: Giải thích các bước giải: a) `4sin \frac{-x}{2} +\sqrt{12}=0` `⇔ sin (\frac{-x}{2})=\frac{-\sqrt{3}}{2}` `⇔ sin (\frac{-x}{2})=sin (\frac{-\pi}{3})` `⇔` \(\left[ \begin{array}{l}\dfrac{-x}{2}=\dfrac{-\pi}{3}+k2\pi\ (k \in \mathbb{Z})\\\dfrac{-x}{2}=\pi-\dfrac{-\pi}{3}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{2\pi}{3}-k4\pi\ (k \in \mathbb{Z})\\x=-\dfrac{8\pi}{3}-k4\pi\ (k \in \mathbb{Z})\end{array} \right.\) b) `\frac{1}{\sqrt{3}} cot (4x-\frac{\pi}{7})=0` `ĐK: x \ne k\pi\ (k \in \mathbb{Z})` `⇔ cot (4x-\frac{\pi}{7})=0` `⇔ \frac{1}{tan (4x-\frac{\pi}{7})}=0` (vô lí) Vậy PT vô nghiệm Reply
Đáp án:
Giải thích các bước giải:
a) `4sin \frac{-x}{2} +\sqrt{12}=0`
`⇔ sin (\frac{-x}{2})=\frac{-\sqrt{3}}{2}`
`⇔ sin (\frac{-x}{2})=sin (\frac{-\pi}{3})`
`⇔` \(\left[ \begin{array}{l}\dfrac{-x}{2}=\dfrac{-\pi}{3}+k2\pi\ (k \in \mathbb{Z})\\\dfrac{-x}{2}=\pi-\dfrac{-\pi}{3}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{2\pi}{3}-k4\pi\ (k \in \mathbb{Z})\\x=-\dfrac{8\pi}{3}-k4\pi\ (k \in \mathbb{Z})\end{array} \right.\)
b) `\frac{1}{\sqrt{3}} cot (4x-\frac{\pi}{7})=0`
`ĐK: x \ne k\pi\ (k \in \mathbb{Z})`
`⇔ cot (4x-\frac{\pi}{7})=0`
`⇔ \frac{1}{tan (4x-\frac{\pi}{7})}=0` (vô lí)
Vậy PT vô nghiệm