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?
 A: I agree with your answer, indeed note that
$$2\tan^2x - 5\sec x = 1\iff2(1-\cos^2x)-5\cos x=\cos^2x\\\iff3\cos^2x+5\cos x-2=0$$
and
$$3t^2+5t-2=0\implies t=\frac{-5\pm\sqrt{25+24}}{6}\implies t=\frac{-5+\sqrt{47}}{6}= \frac13$$
thus we have 2 solution on the interval $[0,2\pi]$ and notably one in the interval $[0,\pi/2]$ and the other in $[3\pi/2,2\pi]$.
Therefore, since the function is periodic with period $2\pi$ we have that $n=13$.
Note that for $x\in\left[0,\frac{13\pi}{2}\right]$ the expression is not defined when $x=\frac{\pi}2+k\pi$ thus this points should be excluded by the solution.
A: simplifying the given equation we get
$$-5\cos(x)^2-5\cos(x)+4=0$$
with $$\cos(x)=t$$ we get the quadratic equation
$$-5t^2-5t+4=0$$
solving this we get
$$t_{1,2}=-\frac{1}{2}\pm\frac{\sqrt{105}}{10}$$
can you finish?
A: 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

