JEE MAIN - Mathematics (2025 - 7th April Evening Shift - No. 13)

The number of solutions of the equation

$ \cos 2\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 :
5
7
6
9

Explanation

$$\begin{aligned} &\begin{aligned} & \cos 2 \theta \cos \frac{\theta}{2}+\cos \frac{5 \theta}{2}=2 \cos ^3 \frac{5 \theta}{2} \\ & \frac{1}{2}\left(2 \cos 2 \theta \cos \frac{\theta}{2}\right)+\cos \frac{50}{2} \\ & =\frac{1}{2}\left(\cos \frac{15 \theta}{2}+3 \cos \frac{5 \theta}{2}\right) \end{aligned}\\ &\text { or solving }\\ &\begin{aligned} & \cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2} \\ & \cos \frac{15 \theta}{2}-\cos \frac{3 \theta}{2}=0 \\ & 2 \sin 30 \sin \frac{9 \theta}{2}=0 \end{aligned}\\ &3 \theta=\mathrm{n} \pi \quad \text { or } \frac{9 \theta}{2}=\mathrm{m} \pi\\ &\begin{aligned} & \theta=\frac{\mathrm{n} \pi}{3} \quad \theta=\frac{2 \mathrm{~m} \pi}{9} \\ & \theta=\left\{-\frac{\pi}{2}, \frac{\pi}{3}, 0\right\} \\ & \theta=\left\{-\frac{4 \pi}{9}, \frac{-2 \pi}{9}, \frac{4 \pi}{9}, \frac{2 \pi}{9}\right\} \end{aligned} \end{aligned}$$

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