<s>[INST] If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is? If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is?
My attempt:
Solving the above quadratic equation, we get $\cos x = \frac{1}{3}$
The general solution of the equation is given by $\cos x = 2n\pi \pm \cos^{-1}\frac{1}{3}$
For having $7$ distinct solutions, $n$ can have value = 0,1,2,3
So, from here we can conclude that $n$ is anything but greater than $6$. So, according to the options given in the questions, the greatest value of $n$ should be $13$. But the answer given is $14$. Can anyone justify?
 [/INST] By replacing $\tan^2 x=\sec^2x-1$ we have$$4\sec^2x-5\sec x=5$$which yields to $$\sec x=\dfrac{5\pm\sqrt{25+80}}{8}=\dfrac{5\pm\sqrt{105}}{8}$$where only $\sec x=\dfrac{5+\sqrt{105}}{8}$ is acceptable. Also the equation $\sec x=p>1$ has exactly two roots in $[0,2\pi)$ so for having 7 distinct roots we must be in the interval $[0,3\times 2\pi+\pi=14\dfrac{\pi}{2})$ which yields to $n<14$ or $n=13$. An illusion is as following

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