JEE Main 2026 — Mathematics Trigonometry

Given equation: $\cos\theta \cos\dfrac{5\theta}{2} = \cos 7\theta \cos\dfrac{7\theta}{2}$cosθcos25θ​=cos7θcos27θ​

Multiplying both sides by $2$2:

$$2\cos\theta \cos\dfrac{5\theta}{2} = 2\cos 7\theta \cos\dfrac{7\theta}{2}$$2cosθcos25θ​=2cos7θcos27θ​

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

$$\cos\left(\theta + \dfrac{5\theta}{2}\right) + \cos\left(\theta - \dfrac{5\theta}{2}\right) = \cos\left(7\theta + \dfrac{7\theta}{2}\right) + \cos\left(7\theta - \dfrac{7\theta}{2}\right)$$cos(θ+25θ​)+cos(θ−25θ​)=cos(7θ+27θ​)+cos(7θ−27θ​)

$$\cos\left(\dfrac{7\theta}{2}\right) + \cos\left(\dfrac{-3\theta}{2}\right) = \cos\left(\dfrac{21\theta}{2}\right) + \cos\left(\dfrac{7\theta}{2}\right)$$cos(27θ​)+cos(2−3θ​)=cos(221θ​)+cos(27θ​)

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

$$\cos\left(\dfrac{7\theta}{2}\right) + \cos\left(\dfrac{3\theta}{2}\right) = \cos\left(\dfrac{21\theta}{2}\right) + \cos\left(\dfrac{7\theta}{2}\right)$$cos(27θ​)+cos(23θ​)=cos(221θ​)+cos(27θ​)

$$\cos\left(\dfrac{3\theta}{2}\right) = \cos\left(\dfrac{21\theta}{2}\right)$$cos(23θ​)=cos(221θ​)

$$\cos\left(\dfrac{21\theta}{2}\right) - \cos\left(\dfrac{3\theta}{2}\right) = 0$$cos(221θ​)−cos(23θ​)=0

Using the identity $\cos C - \cos D = -2\sin\left(\dfrac{C+D}{2}\right)\sin\left(\dfrac{C-D}{2}\right)$cosC−cosD=−2sin(2C+D​)sin(2C−D​):

$$-2\sin\left(\dfrac{\dfrac{21\theta}{2} + \dfrac{3\theta}{2}}{2}\right)\sin\left(\dfrac{\dfrac{21\theta}{2} - \dfrac{3\theta}{2}}{2}\right) = 0$$−2sin​2221θ​+23θ​​​sin​2221θ​−23θ​​​=0

$$-2\sin(6\theta)\sin\left(\dfrac{9\theta}{2}\right) = 0$$−2sin(6θ)sin(29θ​)=0

This gives $\sin(6\theta) = 0$sin(6θ)=0 or $\sin\left(\dfrac{9\theta}{2}\right) = 0$sin(29θ​)=0.

Case 1: $\sin(6\theta) = 0$sin(6θ)=0

$6\theta = n\pi \Rightarrow \theta = \dfrac{n\pi}{6}$6θ=nπ⇒θ=6nπ​ for $n \in \mathbb{Z}$n∈Z.

Since $\theta \in [-\pi, \pi]$θ∈[−π,π], we have $-\pi \le \dfrac{n\pi}{6} \le \pi \Rightarrow -6 \le n \le 6$−π≤6nπ​≤π⇒−6≤n≤6.

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

This gives $13$13 solutions.

Case 2: $\sin\left(\dfrac{9\theta}{2}\right) = 0$sin(29θ​)=0

$\dfrac{9\theta}{2} = m\pi \Rightarrow \theta = \dfrac{2m\pi}{9}$29θ​=mπ⇒θ=92mπ​ for $m \in \mathbb{Z}$m∈Z.

Since $\theta \in [-\pi, \pi]$θ∈[−π,π], we have $-\pi \le \dfrac{2m\pi}{9} \le \pi \Rightarrow -4.5 \le m \le 4.5$−π≤92mπ​≤π⇒−4.5≤m≤4.5.

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

This gives $9$9 solutions.

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

$$\dfrac{n\pi}{6} = \dfrac{2m\pi}{9} \Rightarrow 3n = 4m$$6nπ​=92mπ​⇒3n=4m

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

Thus, there are $3$3 common solutions.

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

Answer: $19$19