1. Rewrite the equation in terms of $\sec\theta$

Given $2\tan^2\theta-5\sec\theta=1.$

Using $\tan^2\theta=\sec^2\theta-1,$ we get $2(\sec^2\theta-1)-5\sec\theta=1.$

So, $2\sec^2\theta-2-5\sec\theta=1$ $2\sec^2\theta-5\sec\theta-3=0.$

Let $x=\sec\theta$. Then $2x^2-5x-3=0.$

Factorizing, $2x^2-6x+x-3=0$ $2x(x-3)+1(x-3)=0$ $(2x+1)(x-3)=0.$

Hence, $\sec\theta=3 \quad \text{or} \quad \sec\theta=-\frac12.$

But $\sec\theta=-\frac12$ is impossible since $|\sec\theta|\ge 1$ for real $\theta$.

So the equation reduces to $\sec\theta=3 \iff \cos\theta=\frac13.$

2. Count solutions of $\cos\theta=\frac13$

Let $\alpha=\cos^{-1}\left(\frac13\right),$ where $0<\alpha<\frac\pi2$.

Then all solutions are $\theta=2m\pi\pm \alpha, \quad m\in\mathbb Z.$

In each full interval of length $2\pi$, there are exactly 2 solutions.

Now we count solutions in $\left[0,\frac{n\pi}{2}\right].$

Let $L=\frac{n\pi}{2}.$

We check the growth of the number of solutions as $L$ increases:

• In $[0,\frac\pi2]$: only $\alpha$ lies inside, so 1 solution.
• In $[0,\pi]$: solutions are $\alpha,\ 2\pi-\alpha$ not included yet, so still 1 solution.
• In $[0,\frac{3\pi}{2}]$: still 1 solution.
• In $[0,2\pi]$: solutions are $\alpha$ and $2\pi-\alpha$, so 2 solutions.

Thus each block of length $2\pi$ contributes 2 solutions, and an additional first-quarter block contributes 1 more solution.

Let us list by $n$ (since endpoint is $n\pi/2$):

• $n=1,2,3$: 1 solution
• $n=4,5,6,7$: 2 solutions
• $n=8,9,10,11$: 4 solutions
• $n=12,13,14,15$: 6 solutions
• $n=16$: interval is $[0,8\pi]$, giving 8 solutions, but maybe 7 occurs earlier in the next quarter after 6.

Let us check after $6\pi$:

In $[0,6\pi]$, solutions are: $\alpha,\ 2\pi-\alpha,\ 2\pi+\alpha,\ 4\pi-\alpha,\ 4\pi+\alpha,\ 6\pi-\alpha$ which are 6 solutions.

Now extend to $\left[0,\frac{13\pi}{2}\right]=[0,6.5\pi]$. The next solution after $6\pi$ is $6\pi+\alpha,$ and since $\alpha<\frac\pi2$, we have $6\pi+\alpha<6\pi+\frac\pi2=\frac{13\pi}{2}.$ So this interval contains exactly 7 solutions.

For $n=12$, upper limit is $6\pi$, giving only 6 solutions. Hence the least such $n$ is $n=13.$

3. Compute the sum

We need $\sum_{k=1}^{13}\frac{k}{2^k}.$

Use the standard formula $\sum_{k=1}^{n} kx^k=\frac{x\left(1-(n+1)x^n+nx^{n+1}\right)}{(1-x)^2}, \quad x\ne 1.$

Put $x=\frac12$ and $n=13$: