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?

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?
