Question

The number of solutions of the equation $|5\tan 2\theta \tan \theta -12\tan \theta |+|3{\sin }^2\theta -\sin 2\theta |=0$ in the interval $[0,2\pi ]$ is

• A. $2$
• B. $3$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

$|5\tan 2\theta \tan \theta -12\tan \theta |+|3{\sin }^2\theta -\sin 2\theta |=0$
$\Rightarrow (5\tan 2\theta -12)\tan \theta =0$
$\Rightarrow \tan \theta =0,\tan 2\theta =\frac{12}{5}$
$\tan 2\theta =\frac{12}{5}$
$\Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{12}{5}$
$\Rightarrow 12{\tan }^2\theta +10\tan \theta -12=0$
$\Rightarrow 6{\tan }^2\theta +5\tan \theta -6=0$
$\Rightarrow (3\tan \theta -2)(2\tan \theta +3)=0$
$\Rightarrow \tan \theta =\frac{2}{3},\frac{-3}{2}$

and $3{\sin }^2\theta -\sin 2\theta =0$
$\Rightarrow 3{\sin }^2\theta -2\sin \theta \cos \theta =0$
$\Rightarrow \sin \theta (3\sin \theta -2\cos \theta )=0$
$\Rightarrow \sin \theta =0,\tan \theta =\frac{2}{3}$
$\therefore$ Common solutions are the solutions corresponding to either $\sin \theta =0$ or $\tan \theta =\frac{2}{3}$

When $\sin \theta =0$
$\Rightarrow \theta =0,\pi ,2\pi$

When $\tan \theta =\frac{2}{3},$
there are two values of $\theta$ in $[0,2\pi ]$

Hence, number of solutions is $5.$