2cot (2x – pi/3) + 1 tìm TXĐ ??? giúp em với ạ 18/08/2021 Bởi Kylie 2cot (2x – pi/3) + 1 tìm TXĐ ??? giúp em với ạ
Đáp án: $\begin{array}{l}2\cot \left( {2x – \frac{\pi }{3}} \right) + 1\\Xác\,định \Leftrightarrow \sin \left( {2x – \frac{\pi }{3}} \right) \ne 0\\ \Rightarrow 2x – \frac{\pi }{3} \ne k\pi \left( {k \in Z} \right)\\ \Rightarrow 2x \ne \frac{\pi }{3} + k\pi \left( {k \in Z} \right)\\ \Rightarrow x \ne \frac{\pi }{6} + \frac{{k\pi }}{2}\left( {k \in Z} \right)\end{array}$ Bình luận
ĐK: $\sin(2x-\dfrac{\pi}{3})\ne 0$ $\Leftrightarrow 2x-\dfrac{\pi}{3}\ne k\pi$ $\Leftrightarrow x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{2}$ $\to D=\mathbb{R}$ \ $\{\dfrac{\pi}{6}+\dfrac{k\pi}{2}\}$ Bình luận
Đáp án:
$\begin{array}{l}
2\cot \left( {2x – \frac{\pi }{3}} \right) + 1\\
Xác\,định \Leftrightarrow \sin \left( {2x – \frac{\pi }{3}} \right) \ne 0\\
\Rightarrow 2x – \frac{\pi }{3} \ne k\pi \left( {k \in Z} \right)\\
\Rightarrow 2x \ne \frac{\pi }{3} + k\pi \left( {k \in Z} \right)\\
\Rightarrow x \ne \frac{\pi }{6} + \frac{{k\pi }}{2}\left( {k \in Z} \right)
\end{array}$
ĐK: $\sin(2x-\dfrac{\pi}{3})\ne 0$
$\Leftrightarrow 2x-\dfrac{\pi}{3}\ne k\pi$
$\Leftrightarrow x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{2}$
$\to D=\mathbb{R}$ \ $\{\dfrac{\pi}{6}+\dfrac{k\pi}{2}\}$