Question

The number of value(s) of $\theta \in [0,2\pi ]$ satisfying the equation $\left({\log }_{\sqrt{5}}\tan \theta \right)\sqrt{{\log }_{\tan \theta }5\sqrt{5}+{\log }_{\sqrt{5}}5\sqrt{5}}=-\sqrt{6}$ is

• A. $0$
• B. $4$
• C. $2$
• D. $5$

Answer 1

The correct option is C $2$

$\left({\log }_{\sqrt{5}}\tan \theta \right)\sqrt{{\log }_{\tan \theta }5\sqrt{5}+{\log }_{\sqrt{5}}5\sqrt{5}}=-\sqrt{6}$
$\Rightarrow {\log }_{\sqrt{5}}\tan \theta \sqrt{\frac{3}{{\log }_{\sqrt{5}}\tan \theta }+3}=-\sqrt{6}$
Put $x={\log }_{\sqrt{5}}\tan \theta$
Then, $x\sqrt{\frac{3}{x}+3}=-\sqrt{6}\dots \left(1\right)$
$\Rightarrow \frac{x\sqrt{x+1}}{\sqrt{x}}=-\sqrt{2}$
$\Rightarrow x^2+x-2=0,x\ne 0$
$\Rightarrow (x+2)(x-1)=0$
$\therefore x=1,-2$

But from $\left(1\right),$ we can conclude $x<0$
$\therefore x=-2$
$\Rightarrow {\log }_{\sqrt{5}}\tan \theta =-2$
$\Rightarrow \tan \theta =\frac{1}{5}$
As $\theta \in [0,2\pi ],$ there are two solutions.