JEE Main 2026 Apr 5 Shift 2, Mathematics Q23: Trigonometric Equations

Use product-to-sum: $\cos A\cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)]$.

LHS $=\frac{1}{2}\left[\cos \frac{3\theta }{2}+\cos \frac{7\theta }{2}\right]$ and RHS $=\frac{1}{2}\left[\cos \frac{7\theta }{2}+\cos \frac{21\theta }{2}\right]$.

Cancelling and simplifying: $\cos \frac{21\theta }{2}-\cos \frac{3\theta }{2}=0$, i.e. $-2\sin 6\theta \sin \frac{9\theta }{2}=0$.

Case 1: $\sin 6\theta =0\Rightarrow \theta =\frac{k\pi }{6}$, $k=-6,-5,\dots ,6$ in $[-\pi ,\pi ]$ — that is 13 solutions.

Case 2: $\sin \frac{9\theta }{2}=0\Rightarrow \theta =\frac{2k\pi }{9}$ in $[-\pi ,\pi ]$, giving $\theta =0,\pm \frac{2\pi }{9},\pm \frac{4\pi }{9},\pm \frac{6\pi }{9},\pm \frac{8\pi }{9}$. Removing $\theta =0$ already counted in Case 1, and $\pm 2\pi /3$ (which is $\pm 4\pi /6$, also in Case 1), the new solutions are $\pm 2\pi /9,\pm 4\pi /9,\pm 8\pi /9$ — 6 more.

$\therefore n(S)=13+6=19$.