Mathematics - Trigonometric Equation Question with Solution | TestHub

To solve $2{\tan }^{2}x-5\sec x=1$, substitute ${\tan }^{2}x={\sec }^{2}x-1$.

This gives $2({\sec }^{2}x-1)-5\sec x=1$, which simplifies to $2{\sec }^{2}x-5\sec x-3=0$.

Factoring, we get $(2\sec x+1)(\sec x-3)=0$.

So, $\sec x=3$ or $\sec x=-1/2$.

Since $\sec x\ge 1$ or $\sec x\le -1$, $\sec x=-1/2$ has no solutions.

Thus, $\sec x=3$, which means $\cos x=1/3$.

Let $\alpha =\arccos (1/3)$. Then $x=2k\pi \pm \alpha$ for $k\in Z$.

We need 7 distinct solutions in $[0,n\pi /2]$.

For $k=0$, $x=\alpha$.

For $k=1$, $x=2\pi -\alpha$ and $x=2\pi +\alpha$.

For $k=2$, $x=4\pi -\alpha$ and $x=4\pi +\alpha$.

For $k=3$, $x=6\pi -\alpha$ and $x=6\pi +\alpha$.

The 7th distinct value is $6\pi -\alpha$.

So, $n\pi /2\ge 6\pi -\alpha$.

Since $\alpha \in (0,\pi /2)$, $6\pi -\alpha <6\pi$.

The next solution would be $6\pi +\alpha$.

To have exactly 7 solutions, $6\pi -\alpha \le n\pi /2<6\pi +\alpha$.

Since $\alpha >0$, we need $n\pi /2$ to be just before $6\pi +\alpha$.

The largest value of $n$ corresponds to $n\pi /2$ being as large as possible without including $6\pi +\alpha$.

The 7 solutions are $\alpha ,2\pi -\alpha ,2\pi +\alpha ,4\pi -\alpha ,4\pi +\alpha ,6\pi -\alpha ,6\pi +\alpha$.

Wait, the solutions are $x=\alpha ,2\pi -\alpha ,2\pi +\alpha ,4\pi -\alpha ,4\pi +\alpha ,6\pi -\alpha$.

The 7th solution is $6\pi -\alpha$.

The solutions are $x=\alpha ,2\pi \pm \alpha ,4\pi \pm \alpha ,6\pi \pm \alpha ,\dots$

In $[0,n\pi /2]$, the solutions are:

$\alpha$ (1st)

$2\pi -\alpha$ (2nd)

$2\pi +\alpha$ (3rd)

$4\pi -\alpha$ (4th)

$4\pi +\alpha$ (5th)

$6\pi -\alpha$ (6th)

$6\pi +\alpha$ (7th)

For exactly 7 distinct values, $6\pi +\alpha \le n\pi /2$.

The next solution would be $8\pi -\alpha$.

So, we need $6\pi +\alpha \le n\pi /2<8\pi -\alpha$.

Since $\alpha =\arccos (1/3)$, $0<\alpha <\pi /2$.

$6\pi +\alpha \le n\pi /2⟹12\pi +2\alpha \le n\pi ⟹12+2\alpha /\pi \le n$.

$n\pi /2<8\pi -\alpha ⟹n\pi <16\pi -2\alpha ⟹n<16-2\alpha /\pi$.

Since $0<2\alpha /\pi <1$, we have $12<12+2\alpha /\pi <13$ and $15<16-2\alpha /\pi <16$.

So, $12+fraction\le n<16-fraction$.

The possible integer values for $n$ are $13,14,15$.

The greatest value of $n$ is 15.

The final answer is $\boxed{15}$.