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}$
