A/tan x +căn 3 B/2 căn 3 tan(2x+π)-2=0 C/tan²(2x)=3 D/[tan(2x+2π/3)-1]+(tan2x)=0

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1. Đáp án:

    Phương trình có các họ nghiệm là:

   $\begin{array}{l}
   a)x =  – \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)\\
   b)x =  – \dfrac{{5\pi }}{{12}} + k\dfrac{\pi }{2}\left( {k \in Z} \right)\\
   c)x =  – \dfrac{\pi }{6} + k\dfrac{\pi }{2}\left( {k \in Z} \right);x = \dfrac{\pi }{6} + k\dfrac{\pi }{2}\left( {k \in Z} \right)\\
   d)x = \dfrac{1}{2}\arctan \left( {\dfrac{{ – 2 + \sqrt 3  + \sqrt {19} }}{{2\sqrt 3 }}} \right) + k\dfrac{\pi }{2}\left( {k \in Z} \right);x = \dfrac{1}{2}\arctan \left( {\dfrac{{ – 2 + \sqrt 3  – \sqrt {19} }}{{2\sqrt 3 }}} \right) + k\dfrac{\pi }{2}\left( {k \in Z} \right)
   \end{array}$

   Giải thích các bước giải:

   $\begin{array}{l}
   a)\tan x + \sqrt 3  = 0\\
    \Leftrightarrow \tan x =  – \sqrt 3 \\
    \Leftrightarrow x =  – \dfrac{\pi }{3} + k\pi \\
   b)2\sqrt 3 \tan \left( {2x + \pi } \right) – 2 = 0\\
    \Leftrightarrow \tan \left( {2x + \pi } \right) = \dfrac{1}{{\sqrt 3 }}\\
    \Leftrightarrow 2x + \pi  = \dfrac{\pi }{6} + k\pi \\
    \Leftrightarrow x =  – \dfrac{{5\pi }}{{12}} + k\dfrac{\pi }{2}\\
   c){\tan ^2}2x = 3\\
    \Leftrightarrow \left[ \begin{array}{l}
   \tan 2x =  – \sqrt 3 \\
   \tan 2x = \sqrt 3 
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   2x =  – \dfrac{\pi }{3} + k\pi \\
   2x = \dfrac{\pi }{3} + k\pi 
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x =  – \dfrac{\pi }{6} + k\dfrac{\pi }{2}\\
   x = \dfrac{\pi }{6} + k\dfrac{\pi }{2}
   \end{array} \right.\\
   d)\left[ {\tan \left( {2x + \dfrac{{2\pi }}{3}} \right) – 1} \right] + \tan 2x = 0\\
    \Leftrightarrow \tan \left( {2x + \dfrac{{2\pi }}{3}} \right) – 1 + \tan 2x = 0\\
    \Leftrightarrow \dfrac{{\tan 2x + \tan \dfrac{{2\pi }}{3}}}{{1 – \tan 2x.\tan \dfrac{{2\pi }}{3}}} – 1 + \tan 2x = 0\\
    \Leftrightarrow \dfrac{{\tan 2x – \sqrt 3 }}{{1 + \tan 2x.\sqrt 3 }} – 1 + \tan 2x = 0\\
    \Leftrightarrow \dfrac{{\tan 2x – \sqrt 3  + \left( {\tan 2x – 1} \right)\left( {1 + \tan 2x.\sqrt 3 } \right)}}{{1 + \tan 2x.\sqrt 3 }} = 0\\
    \Leftrightarrow \sqrt 3 .{\tan ^2}2x + \tan 2x\left( {2 – \sqrt 3 } \right) – 1 – \sqrt 3  = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   \tan 2x = \dfrac{{ – 2 + \sqrt 3  + \sqrt {19} }}{{2\sqrt 3 }}\\
   \tan 2x = \dfrac{{ – 2 + \sqrt 3  – \sqrt {19} }}{{2\sqrt 3 }}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   2x = \arctan \left( {\dfrac{{ – 2 + \sqrt 3  + \sqrt {19} }}{{2\sqrt 3 }}} \right) + k\pi \\
   2x = \arctan \left( {\dfrac{{ – 2 + \sqrt 3  – \sqrt {19} }}{{2\sqrt 3 }}} \right) + k\pi 
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = \dfrac{1}{2}\arctan \left( {\dfrac{{ – 2 + \sqrt 3  + \sqrt {19} }}{{2\sqrt 3 }}} \right) + k\dfrac{\pi }{2}\\
   x = \dfrac{1}{2}\arctan \left( {\dfrac{{ – 2 + \sqrt 3  – \sqrt {19} }}{{2\sqrt 3 }}} \right) + k\dfrac{\pi }{2}
   \end{array} \right.
   \end{array}$

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