Đáp án:
\(TXD:D = R\backslash \left\{ { - \dfrac{\pi }{3} + k\pi ;\dfrac{\pi }{{12}} + \dfrac{{k\pi }}{3};\dfrac{\pi }{4} + \dfrac{{k\pi }}{3}} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:\left\{ \begin{array}{l}
\tan \left( {3x - \dfrac{\pi }{4}} \right) \ne 0\\
\sin \left( {\dfrac{\pi }{3} + x} \right) \ne 0\\
\cos \left( {3x - \dfrac{\pi }{4}} \right) \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3x - \dfrac{\pi }{4} \ne k\pi \\
\dfrac{\pi }{3} + x \ne k\pi \\
3x - \dfrac{\pi }{4} \ne \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3x \ne \dfrac{\pi }{4} + k\pi \\
x \ne - \dfrac{\pi }{3} + k\pi \\
3x \ne \dfrac{{3\pi }}{4} + k\pi
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{3}\\
x \ne - \dfrac{\pi }{3} + k\pi \\
x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{3}
\end{array} \right.\\
\to TXD:D = R\backslash \left\{ { - \dfrac{\pi }{3} + k\pi ;\dfrac{\pi }{{12}} + \dfrac{{k\pi }}{3};\dfrac{\pi }{4} + \dfrac{{k\pi }}{3}} \right\}
\end{array}\)
