JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Trigonometric Equations

$2{\tan }^2\theta -5\sec \theta =1$. Using ${\tan }^2\theta ={\sec }^2\theta -1$:


$\sec \theta =-\frac{1}{2}$ is impossible since $|\sec \theta |\ge 1$, so $\sec \theta =3$, i.e. $\cos \theta =\frac{1}{3}$.

In every interval of length $2\pi$, $\cos \theta =\frac{1}{3}$ has exactly two solutions, ${\theta }_1={\cos }^{-1}\left(\frac{1}{3}\right)\approx 1.231$ and ${\theta }_2=2\pi -{\theta }_1\approx 5.052$, repeating every $2\pi$. The successive solutions are approximately


For exactly 7 solutions in $\left[0,\frac{n\pi }{2}\right]$, the interval must reach the 7th solution ($\approx 20.080$) but not the 8th ($\approx 23.90$). With $n=13$, $\frac{n\pi }{2}=\frac{13\pi }{2}\approx 20.42$, which includes exactly 7 solutions; with $n=12$, $\frac{12\pi }{2}=6\pi \approx 18.85$, which includes only 6. So the least such $n$ is $13$.

Now evaluate $S=\sum _{k=1}^{13}\frac{k}{2^k}$.



Subtracting: