Question

The number of solution(s) of the equation $({\log }_2\cos \theta {)}^2+{\log }_{\frac{4}{\cos \theta }}(16\cos \theta )=2$ in the interval $[0,2\pi )$ is

Answer 1

Let $\cos \theta =t$
Then, $({\log }_{2}t{)}^2+{\log }_{4/t}(16t)=2$
$\Rightarrow ({\log }_{2}t{)}^2+\frac{{\log }_2\left(16t\right)}{{\log }_2\left(\frac{4}{t}\right)}=2$ (Using base change property)
$\Rightarrow ({\log }_{2}t{)}^2+\frac{4+{\log }_{2}t}{2-{\log }_{2}t}=2$

Let ${\log }_{2}t=z$
$z^2+\frac{4+z}{2-z}=2$
$\Rightarrow z(z^2-2z-3)=0$
$\Rightarrow z=0,-1,3$
$\Rightarrow t=1,\frac{1}{2},8$

So, $\cos \theta =\frac{1}{2}$ or $\cos \theta =1$
$\Rightarrow \theta =2n\pi \pm \frac{\pi }{3}$ or $\theta =2m\pi ;n,m\in I$
Hence, the number of solutions in $[0,2\pi )$ is $3(\theta =\frac{\pi }{3},\frac{5\pi }{3},0)$