The number of solutions of the equation (lo g _ 2 cos θ) ^ 2 + lo g...

For domain,& $cos \theta > 0 ⇒ \theta \in \left[0, \dfrac{π}{2}\right)\cup \left(\dfrac{3\pi}{2},2\pi\right)<br />Also, if$cos $\theta$> 0 ⇒ $\dfrac{4}{cos \theta} \ne$1$<br />$⇒ { \left(lo{g}_{2}cos \theta\right)}^{2} + \dfrac{lo{g}_{2} \left(16cos \theta\right)}{lo{g}_{2} \left(\dfrac{4}{cos \theta}\right)} =2
$⇒${\lleft(lo{g}_{2}cos\theta)}^{2} + \left(\dfrac{lo{g}_{2}16 + lo{g}_{2}cos\theta}{lo{g}_{2}4 - lo{g}_{2}cos\theta}\right) =$2$
Let,~$lo{g}_{2}cos\theta =t$

$⇒ t^{2} + \left(\dfrac{4 + t}{2 - t}\right)=2$
$⇒ 2t^{2} - t^{3} + 4 + t = 4 - 2t ⇒ t^{3} - 2t^{2} - 3t = 0$
$⇒ t = 0,3, - 1$ .
$t = 0 ⇒ lo{g}_{2}cos\theta =0 ⇒ cos\theta =1 ⇒ \theta =0$
$t = - 1 ⇒ lo{g}_{2}cos\theta = -1 ⇒ cos\theta = \dfrac{1}{2}⇒ \theta = \dfrac{\pi}{3},\dfrac{5\pi}{3}$ .
$t = 3 ⇒ lo{g}_{2}cos\theta =3 ⇒ cos\theta =8$~(not possible)
$⇒$ Number of solutions are~$3$