<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] 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.
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