Giải phương trình $\sqrt 3 \cot (5x - \frac{\pi }{8}) = 0$.
A. $x = \frac{\pi }{8} + k\pi ;k \in \mathbb{Z}$.
B. $x = \frac{\pi }{8} + k\frac{\pi }{5};k \in \mathbb{Z}$.
C. $x = \frac{\pi }{8} + k\frac{\pi }{4};k \in \mathbb{Z}$.
D. $x = \frac{\pi }{8} + k\frac{\pi }{2};k \in \mathbb{Z}$.
A. $x = \frac{\pi }{8} + k\pi ;k \in \mathbb{Z}$.
B. $x = \frac{\pi }{8} + k\frac{\pi }{5};k \in \mathbb{Z}$.
C. $x = \frac{\pi }{8} + k\frac{\pi }{4};k \in \mathbb{Z}$.
D. $x = \frac{\pi }{8} + k\frac{\pi }{2};k \in \mathbb{Z}$.
Chọn B.
Ta có $\sqrt 3 \cot \left( {5x - \frac{\pi }{8}} \right) = 0 \Leftrightarrow \cot \left( {5x - \frac{\pi }{8}} \right) = 0 \Leftrightarrow \cos \left( {5x - \frac{\pi }{8}} \right) = 0$
$ \Leftrightarrow 5x - \frac{\pi }{8} = \frac{\pi }{2} + k\pi \Leftrightarrow x = \frac{\pi }{8} + \frac{{k\pi }}{5};k \in \mathbb{Z}$.
Ta có $\sqrt 3 \cot \left( {5x - \frac{\pi }{8}} \right) = 0 \Leftrightarrow \cot \left( {5x - \frac{\pi }{8}} \right) = 0 \Leftrightarrow \cos \left( {5x - \frac{\pi }{8}} \right) = 0$
$ \Leftrightarrow 5x - \frac{\pi }{8} = \frac{\pi }{2} + k\pi \Leftrightarrow x = \frac{\pi }{8} + \frac{{k\pi }}{5};k \in \mathbb{Z}$.