Question

Let $A=\begin{bmatrix} 1 & tan x \\ -tan⁡x & 1 \end{bmatrix}$ and $B=A^{T}A^{- 2}$ . If $f\left(x\right)=det\left(a d j B\right)$ , then number of solution(s) of the equation $10f\left(x\right)-x=0$ is (are)

Answer 1

$f(x)=\operatorname{det}(\operatorname{adj}(B))=|\operatorname{adj}(B)|=|B|$
$=\left|A^{T} \| A^{-2}\right|=|A| \frac{1}{\mid A}$
$=\frac{1}{\left|\right. A \left|\right.}=\frac{1}{s e c^{2} x}=cos^{2}x$

$10cos^{2}x-x=0$