JEE Main 2024Trigonometric EquationsTrigonometric EquationsMediumQuestionThe number of solutions of the equation cos2θcosθ2+cos5θ2=2cos35θ2\cos 2\theta \cos \frac{\theta}{2} + \cos \frac{5\theta}{2} = 2\cos^3 \frac{5\theta}{2}cos2θcos2θ+cos25θ=2cos325θ in [−π2,π2]\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right][−2π,2π] is :OptionsA5B7C6D9Check AnswerHide SolutionSolutioncos2θcosθ2+cos5θ2=2cos35θ212(2cos2θcosθ2)+cos502=12(cos15θ2+3cos5θ2) or solving cos3θ2=cos15θ2cos15θ2−cos3θ2=02sin30sin9θ2=03θ=nπ or 9θ2=mπθ=nπ3θ=2 mπ9θ={−π2,π3,0}θ={−4π9,−2π9,4π9,2π9}\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}cos2θcos2θ+cos25θ=2cos325θ21(2cos2θcos2θ)+cos250=21(cos215θ+3cos25θ) or solving cos23θ=cos215θcos215θ−cos23θ=02sin30sin29θ=03θ=nπ or 29θ=mπθ=3nπθ=92 mπθ={−2π,3π,0}θ={−94π,9−2π,94π,92π}