JEE Main 2025 Apr 7 Shift 2, Mathematics Q9: Trigonometric Equations

Using $2{\cos }^{3}x=\frac{1}{2}\left(\cos 3x+3\cos x\right)$ with $x=\frac{5\theta }{2}$, the right side becomes $\frac{1}{2}\cos \frac{15\theta }{2}+\frac{3}{2}\cos \frac{5\theta }{2}$.

Also, $\cos 2\theta \cos \frac{\theta }{2}=\frac{1}{2}\left[\cos \frac{5\theta }{2}+\cos \frac{3\theta }{2}\right]$ by the product-to-sum identity.

Substituting both into the equation and cancelling the common $\frac{1}{2}\cos \frac{5\theta }{2}$ term from each side reduces it to $\cos \frac{3\theta }{2}=\cos \frac{15\theta }{2}$.

This gives $\frac{15\theta }{2}=2n\pi \pm \frac{3\theta }{2}$, leading to two families: $\theta =\frac{n\pi }{3}$ or $\theta =\frac{2m\pi }{9}$.

Within $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, the first family gives $\theta \in \left\{-\frac{\pi }{3},0,\frac{\pi }{3}\right\}$ and the second gives $\theta \in \left\{-\frac{4\pi }{9},-\frac{2\pi }{9},0,\frac{2\pi }{9},\frac{4\pi }{9}\right\}$.

Combining and removing the shared value $\theta =0$ gives 7 distinct solutions.