Đáp án:
\(\begin{array}{l}
f,\\
D = R\backslash \left\{ {\dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}|k \in Z} \right\}\\
g,\\
D = R\backslash \left\{ {\dfrac{{2\pi }}{3} + k2\pi |k \in Z} \right\}\\
h,\\
D = R\backslash \left\{ {\dfrac{\pi }{4} + k\pi |k \in Z} \right\}
\end{array}\)
Giải thích các bước giải:
Hàm số đã cho xác định khi và chỉ khi:
\(\begin{array}{l}
f,\\
\cos \left( {4x + \dfrac{\pi }{6}} \right) \ne 0\\
\Leftrightarrow 4x + \dfrac{\pi }{6} \ne \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow 4x \ne \dfrac{\pi }{3} + k\pi \\
\Leftrightarrow x \ne \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}\,\,\,\,\left( {k \in Z} \right)\\
\Rightarrow TXD:\,\,\,\,D = R\backslash \left\{ {\dfrac{\pi }{{12}} + \dfrac{{k\pi }}{4}|k \in Z} \right\}\\
g,\\
\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{3}} \right) \ne 0\\
\Leftrightarrow \dfrac{x}{2} - \dfrac{\pi }{3} \ne k\pi \\
\Leftrightarrow \dfrac{x}{2} \ne \dfrac{\pi }{3} + k\pi \\
\Leftrightarrow x \ne \dfrac{{2\pi }}{3} + k2\pi \,\,\,\,\,\left( {k \in Z} \right)\\
\Rightarrow TXD:\,\,\,\,D = R\backslash \left\{ {\dfrac{{2\pi }}{3} + k2\pi |k \in Z} \right\}\\
h,\\
\sin 2x - 1 \ne 0\\
\Leftrightarrow \sin 2x \ne 1\\
\Leftrightarrow 2x \ne \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x \ne \dfrac{\pi }{4} + k\pi \,\,\,\,\,\left( {k \in Z} \right)\\
\Rightarrow TXD:\,\,\,\,\,D = R\backslash \left\{ {\dfrac{\pi }{4} + k\pi |k \in Z} \right\}
\end{array}\)
