[Math] If $2\tan^2 x – 5\sec x – 1 = 0$ has 7 roots in $[ 0,\frac{n\pi}{2}]$ then the greatest value of $n$ is

Question

What I did so far :

$2\tan^2x – 5\sec x – 1 = 0$

$\Rightarrow 2\sec^2x – 2 – 5\sec x -1 = 0$

$\Rightarrow ( 2\sec x + 1 ) ( \sec x – 3 ) = 0$

$\Rightarrow \sec x = 3$

I need to get the solution in general form and find the value of $n$ .
This is a high school math problem that needs to be done without a calculator . What is the correct way to solve it ?

Answer 1

We have $\cos x=\frac 13.$

For $n=1,5,9,13,\cdots$, the number of different roots in $[0,\frac{n\pi}{2}]$ is $1,3,5,7,\cdots$ respectively.

So, the greatest value of $n$ is $13+2=\color{red}{15}$.