JEE Main 2025 — Trigonometric Equations Question with Solution
JEE Main 2025 (7 Apr Shift 2)
Question
The number of solutions of the equation
in is :
in is :
Choose an option
Show full solutionCorrect option: A
Correct answer
A7
Step-by-step explanation
$\begin{aligned}
& \cos 2 \theta \cos \frac{\theta}{2}+\cos \frac{5 \theta}{2}=2 \cos ^3 \frac{5 \theta}{2} \\ & \frac{1}{2}\left(2 \cos 2 \theta \cos \frac{\theta}{2}\right)+\cos \frac{50}{2} \\ & =\frac{1}{2}\left(\cos \frac{15 \theta}{2}+3 \cos \frac{5 \theta}{2}\right)
\end{aligned}\begin{aligned}
& \cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2} \\ & \cos \frac{15 \theta}{2}-\cos \frac{3 \theta}{2}=0 \\ & 2 \sin 30 \sin \frac{9 \theta}{2}=0
\end{aligned}3 \theta=\mathrm{n} \pi \text { or } \frac{9 \theta}{2}=\mathrm{m} \pi\begin{aligned}
& \theta=\frac{\mathrm{n} \pi}{3} \quad \theta=\frac{2 \mathrm{~m} \pi}{9} \\ & \theta=\left\{-\frac{\pi}{2}, \frac{\pi}{3}, 0\right\} \\ & \theta=\left\{-\frac{4 \pi}{9}, \frac{-2 \pi}{9}, \frac{4 \pi}{9}, \frac{2 \pi}{9}\right\}
\end{aligned}$
& \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}\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 \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}$
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This is a previous-year question from JEE Main 2025, covering the Trigonometric Equations chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.