JEE Main 2026 — Trigonometric Equations Question with Solution

Given equation: $\cos \theta \cos \frac{5\theta }{2}=\cos 7\theta \cos \frac{7\theta }{2}$

Multiplying both sides by $2$:

$2\cos \theta \cos \frac{5\theta }{2}=2\cos 7\theta \cos \frac{7\theta }{2}$

Using the identity $2\cos A\cos B=\cos (A+B)+\cos (A-B)$:

$\cos \left(\theta +\frac{5\theta }{2}\right)+\cos \left(\theta -\frac{5\theta }{2}\right)=\cos \left(7\theta +\frac{7\theta }{2}\right)+\cos \left(7\theta -\frac{7\theta }{2}\right)$

$\cos \left(\frac{7\theta }{2}\right)+\cos \left(\frac{-3\theta }{2}\right)=\cos \left(\frac{21\theta }{2}\right)+\cos \left(\frac{7\theta }{2}\right)$

Since $\cos (-x)=\cos x$, we get:

$\cos \left(\frac{7\theta }{2}\right)+\cos \left(\frac{3\theta }{2}\right)=\cos \left(\frac{21\theta }{2}\right)+\cos \left(\frac{7\theta }{2}\right)$

$\cos \left(\frac{3\theta }{2}\right)=\cos \left(\frac{21\theta }{2}\right)$

$\cos \left(\frac{21\theta }{2}\right)-\cos \left(\frac{3\theta }{2}\right)=0$

Using the identity $\cos C-\cos D=-2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$:

$-2\sin \left(\frac{\frac{21\theta }{2}+\frac{3\theta }{2}}{2}\right)\sin \left(\frac{\frac{21\theta }{2}-\frac{3\theta }{2}}{2}\right)=0$

$-2\sin (6\theta )\sin \left(\frac{9\theta }{2}\right)=0$

This gives $\sin (6\theta )=0$ or $\sin \left(\frac{9\theta }{2}\right)=0$.

Case 1: $\sin (6\theta )=0$

$6\theta =n\pi \Rightarrow \theta =\frac{n\pi }{6}$ for $n\in \mathbb{Z}$.

Since $\theta \in [-\pi ,\pi ]$, we have $-\pi \le \frac{n\pi }{6}\le \pi \Rightarrow -6\le n\le 6$.

The possible values for $n$ are $\{-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6\}$.

This gives $13$ solutions.

Case 2: $\sin \left(\frac{9\theta }{2}\right)=0$

$\frac{9\theta }{2}=m\pi \Rightarrow \theta =\frac{2m\pi }{9}$ for $m\in \mathbb{Z}$.

Since $\theta \in [-\pi ,\pi ]$, we have $-\pi \le \frac{2m\pi }{9}\le \pi \Rightarrow -4.5\le m\le 4.5$.

The possible values for $m$ are $\{-4,-3,-2,-1,0,1,2,3,4\}$.

This gives $9$ solutions.

To find common solutions, we equate the two sets of solutions:

$\frac{n\pi }{6}=\frac{2m\pi }{9}\Rightarrow 3n=4m$

For $m\in \{-4,-3,-2,-1,0,1,2,3,4\}$, $m$ must be a multiple of $3$. The valid values for $m$ are $-3,0,3$, which correspond to $n=-4,0,4$.

Thus, there are $3$ common solutions.

Total number of unique solutions $n(S)=13+9-3=19$.

Answer: $19$