JEE Main 2025 (Online) 7th April Evening ShiftQuestion 1
The number of solutions of the equation $\cos2\theta\cos\frac{\theta}{2}+\cos\frac{5\theta}{2}=2\cos^{3}\frac{5\theta}{2}$ in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ is :
Trigonometric Ratio & Identities
Solution
Concept. Write $2\cos^3\frac{5\theta}2-\cos\frac{5\theta}2$ as a product; then both sides become sums and one term cancels. Formulas used. (1) $2\cos^2u-1=\cos2u$(2) $2\cos A\cos B=\cos(A+B)+\cos(A-B)$, and $\cos x=\cos y\iff x=2k\pi\pm y$Step 1 - Rewrite. By (1), $2\cos^3\frac{5\theta}{2}-\cos\frac{5\theta}{2}=\cos\frac{5\theta}{2}\cos5\theta$. The equation is $$\cos2\theta\cos\frac\theta2=\cos\frac{5\theta}{2}\cos5\theta.$$ Step 2 - Product to sum. By (2): $$\cos\frac{5\theta}{2}+\cos\frac{3\theta}{2}=\cos\frac{15\theta}{2}+\cos\frac{5\theta}{2}\ \Longrightarrow\ \cos\frac{15\theta}{2}=\cos\frac{3\theta}{2}.$$ Step 3 - Solve. $\frac{15\theta}{2}=2k\pi\pm\frac{3\theta}{2}$ gives $\theta=\frac{k\pi}{3}$ or $\theta=\frac{2k\pi}{9}$. Step 4 - Count in $\left[-\frac\pi2,\frac\pi2\right]$. $\theta=0,\pm\frac\pi3$ and $\theta=0,\pm\frac{2\pi}9,\pm\frac{4\pi}9$ ($\frac{4\pi}{9}<\frac\pi2$). Together: $0,\pm\frac{2\pi}9,\pm\frac\pi3,\pm\frac{4\pi}9$, i.e. $7$ solutions. Answer: (C)