The number of distinct real roots of the folllowing determinant

Question

The number of distinct real roots of determinant $$ \begin{vmatrix} \csc x & \sec x & \sec x \\ \sec x & \csc x & \sec x \\ \sec x & \sec x & \csc x \\ \end{vmatrix} =0$$ lies in the interval $\frac{-\pi}{4} \le x \le \frac{\pi}{4}$ is
(a) $1$
(b) $2$
(c) $3$
(d) $0$

I tried to solve the determinant and solved it till I got:
$$(\csc x + 2\sec x)(\csc x - \sec x)^2=0$$

How do I get the final answer?
Please offer your assistance,
Thank you

Answer 1

We need to solve $\cos(x) + 2\sin(x) = 0$ and $\cos(x) - \sin(x) = 0$. The first equation is solved by setting $u = \cos(x)$ and solving for $u$, i.e. \begin{align*} u+2\sqrt{1-u^2} &= 0\\ u &= -2\sqrt{1-u^2} \\ u^2 &= 4(1-u^2) \\ 5u^2 &= 4 \end{align*} and checking which values of $\cos^{-1}(\pm 2/\sqrt{5})$ lie in the specified interval. For the second equation, only $x = \pi/4$ works.

Answer 2

For the root of given determinant, you need either $\csc x+2\sec x=0$ or $\csc x-\sec x=0$.

i.e. $\tan x=-1/2$ or $tan x=1$

As $\tan x$ is bijective in $(-\pi/4,\pi/4)$ so both conditions give you a unique value of $x$ in prescribed interval. Hence the correct answer is $2$.