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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: $\sqrt{3}tan^{2}x + 4tanx + \sqrt{3} = 0$ $3cotx + tanx + 4 = 0$

Tháng Tám 15, 2022 Toán học 2 Comments 0 views

Toán Lớp 11: $\sqrt{3}tan^{2}x + 4tanx + \sqrt{3} = 0$
$3cotx + tanx + 4 = 0$

Comments ( 2 )

  1. haianhtu97
    haianhtu97
    15 Tháng Tám, 2022 at 12:07 sáng
    Reply
    chúc bn hc tốt vote mk 5* và ctlhn nhé
     

    toan-lop-11-sqrt-3-tan-2-4tan-sqrt-3-0-3cot-tan-4-0

  2. ngochuong2k2
    ngochuong2k2
    15 Tháng Tám, 2022 at 12:06 sáng
    Reply
    Giải đáp:
     
    Lời giải và giải thích chi tiết:
     \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})}
    ——————————–
    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
    ⇔ \(\left[ \begin{array}{l} \tan\ x=-1\\\tan\ x=-3\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l} \tan\ x=\tan\ (-\dfrac{\pi}{4})\\x=\arctan\ (-3)+k\pi\ (k \in \mathbb{Z})\end{array} \right.\) 
    ⇔ \(\left[ \begin{array}{l} x=-\dfrac{\pi}{4}+k\pi\ (k \in \mathbb{Z})\\x=\arctan\ (-3)+k\pi\ (k \in \mathbb{Z})\end{array} \right.\) 

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222-9+11+12:2*14+14 = ? ( )

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