JEE Main 2025 — 7 April, Evening Shift — Question 22

$2(\cos 2 \theta) \cdot\left(\cos \frac{\theta}{2}\right)+2 \cos \frac{5 \theta}{2}=4 \cos ^{3}\left(\frac{5 \theta}{2}\right)$

$$\Rightarrow \cos \left(\frac{5 \theta}{2}\right)+\cos \frac{3 \theta}{2}+2 \cos \left(\frac{5 \theta}{2}\right)$$ $$=\left(\cos \frac{15 \theta}{2}+3 \cos \frac{5 \theta}{2}\right)$$

$\Rightarrow \cos \left(\frac{3 \theta}{2}\right)+\cos \left(\frac{15 \theta}{2}\right)$

$\Rightarrow \cos \left(\frac{3 \theta}{2}\right)-\cos \frac{15 \theta}{2}=0$

$\Rightarrow 2 \sin \left(\frac{9 \theta}{2}\right) \sin \left(\frac{6 \theta}{2}\right)=0,3 \theta=2 n \pi$

$\therefore \quad \frac{9 \theta}{2}=\eta \pi \rightarrow \theta=\frac{2 \eta \pi}{9}$

$\Rightarrow \quad \theta=\frac{2 \eta \pi}{3}$

$\therefore \quad \theta=-\frac{4 \pi}{9},-\frac{3 \pi}{9},-\frac{2 \pi}{9}, 0, \frac{2 \pi}{9}, \frac{3 \pi}{9}, \frac{4 \pi}{9}$