Đáp án:
Giải thích các bước giải:
`\sqrt{3}tan^2 x+4tan\ x+\sqrt{3}=0`
`⇔ (tan\ x+\sqrt{3})(\sqrt{3}tan\ x+1)=0`
`⇔` \(\left[ \begin{array}{l} \tan\ x+\sqrt{3}=0\\\sqrt{3}\tan\ x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l} \tan\ x=-\sqrt{3}\\\tan\ x=-\dfrac{1}{\sqrt{3}}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l} \tan\ x=\tan\ (-\dfrac{\pi}{3})\\\tan\ x=\tan\ (-\dfrac{\pi}{6})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l} x=-\dfrac{\pi}{3}+k\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{6}+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={-\frac{\pi}{3}+k\pi\ (k \in \mathbb{Z});-\frac{\pi}{6}+k\pi\ (k \in \mathbb{Z})}`
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`3cot\ x+tan\ x+4=0`
`⇔ \frac{3}{tan\ x}+tan\ x+4=0`
ĐK: `x \ne k\frac{\pi}{2}\ (k \in \mathbb{Z})`
`⇔ 3+tan^2 x+4tan\ x=0`
`⇔ (tan\ x+1)(tan\ x+3)=0`
`⇔` \(\left[ \begin{array}{l} \tan\ x=-1\\\tan\ x=-3\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l} \tan\ x=\tan\ (-\dfrac{\pi}{4})\\x=\arctan\ (-3)+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l} x=-\dfrac{\pi}{4}+k\pi\ (k \in \mathbb{Z})\\x=\arctan\ (-3)+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)
