Phương trình: $\tan \left( {\frac{\pi }{2} - x} \right) + 2\tan \left( {2x + \frac{\pi }{2}} \right) = 1$ có nghiệm là:
A. $x = \frac{\pi }{4} + k2\pi \left( {k \in \mathbb{Z}} \right)$
B. $x = \frac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right)$
C. $x = \frac{\pi }{4} + k\frac{\pi }{2}\left( {k \in \mathbb{Z}} \right)$
D. $x = \pm \frac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right)$
A. $x = \frac{\pi }{4} + k2\pi \left( {k \in \mathbb{Z}} \right)$
B. $x = \frac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right)$
C. $x = \frac{\pi }{4} + k\frac{\pi }{2}\left( {k \in \mathbb{Z}} \right)$
D. $x = \pm \frac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right)$
Chọn B.
Ta có : $\tan \left( {\frac{\pi }{2} - x} \right) + 2\tan \left( {2x + \frac{\pi }{2}} \right) = 1$
$ \Leftrightarrow \cot x - 2\cot 2x = 1$$ \Leftrightarrow \cot x - 2\frac{{1 - {{\tan }^2}x}}{{2\tan x}} = 1$
$ \Leftrightarrow \cot x - (\cot x - \tan x) = 1$
$ \Leftrightarrow \tan x = 1$
$ \Leftrightarrow x = \frac{\pi }{4} + k\pi \;\left( {k \in \mathbb{Z}} \right)$
Ta có : $\tan \left( {\frac{\pi }{2} - x} \right) + 2\tan \left( {2x + \frac{\pi }{2}} \right) = 1$
$ \Leftrightarrow \cot x - 2\cot 2x = 1$$ \Leftrightarrow \cot x - 2\frac{{1 - {{\tan }^2}x}}{{2\tan x}} = 1$
$ \Leftrightarrow \cot x - (\cot x - \tan x) = 1$
$ \Leftrightarrow \tan x = 1$
$ \Leftrightarrow x = \frac{\pi }{4} + k\pi \;\left( {k \in \mathbb{Z}} \right)$
