Question

If $x\in [0,2\pi ]$ and ${\log }_2\left(\tan x\right)+{\log }_2\left(\tan 2x\right)=0,$ then $x$ can be

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{6}$
• C. $\frac{7\pi }{6}$
• D. $\frac{5\pi }{6}$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{\pi }{6}$
C $\frac{7\pi }{6}$

${\log }_2\left(\tan x\right)+{\log }_2\left(\tan 2x\right)=0$
$\log$ function is defined when $\tan x,\tan 2x>0$
$\Rightarrow n\pi <x<n\pi +\frac{\pi }{4},n\in Z$
Now, ${\log }_2(\tan x\cdot \tan 2x)=0$
$\Rightarrow \frac{2{\tan }^{2}x}{1-{\tan }^{2}x}=1$
$\Rightarrow {\tan }^{2}x=\frac{1}{3}$
$\Rightarrow \tan x=\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\text{(rejected)}$
$\Rightarrow x=\frac{\pi }{6},\frac{7\pi }{6}$