Đáp án:$\left[ \begin{array}{l}
x = 1 + \dfrac{\pi }{3} + k\pi \\
x = 1 - \dfrac{\pi }{3} - k\pi
\end{array} \right.\left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
\tan \left| {x - {5^0}} \right| = \sqrt 3 \\
\Leftrightarrow \tan \left| {x - 1} \right| = \sqrt 3 \\
\Leftrightarrow \left| {x - 1} \right| = \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1 + \dfrac{\pi }{3} + k\pi \\
x = 1 - \dfrac{\pi }{3} - k\pi
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,\left[ \begin{array}{l}
x = 1 + \dfrac{\pi }{3} + k\pi \\
x = 1 - \dfrac{\pi }{3} - k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
