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 :
$ \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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