JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Trigonometric Equations

By product-to-sum, $\cos \theta \cos \frac{5\theta }{2}=\frac{1}{2}\left(\cos \frac{7\theta }{2}+\cos \frac{3\theta }{2}\right)$ and $\cos 7\theta \cos \frac{7\theta }{2}=\frac{1}{2}\left(\cos \frac{21\theta }{2}+\cos \frac{7\theta }{2}\right)$. Equating cancels $\cos \frac{7\theta }{2}$ and leaves $\cos \frac{3\theta }{2}=\cos \frac{21\theta }{2}$, i.e. $\cos \frac{21\theta }{2}-\cos \frac{3\theta }{2}=-2\sin 6\theta \sin \frac{9\theta }{2}=0$. From $\sin 6\theta =0$, $\theta =\frac{k\pi }{6}$ gives $13$ values in $[-\pi ,\pi ]$ $(k=-6,\dots ,6)$; from $\sin \frac{9\theta }{2}=0$, $\theta =\frac{2m\pi }{9}$ gives $9$ values $(m=-4,\dots ,4)$. The two lists share $\theta =0,\pm \frac{2\pi }{3}$, three values in common, so the number of distinct solutions is $13+9-3=19$. Thus $n(S)=19$.