JEE Main 2020Trigonometric EquationsTrigonometric EquationsHardQuestionIf m and n respectively are the numbers of positive and negative values of θ\thetaθ in the interval [−π,π][-\pi,\pi][−π,π] that satisfy the equation cos2θcosθ2=cos3θcos9θ2\cos 2\theta \cos {\theta \over 2} = \cos 3\theta \cos {{9\theta } \over 2}cos2θcos2θ=cos3θcos29θ, then mn is equal to ____________.Answer: 2Hide SolutionSolution2cos2θcosθ2=2cos3θcos9θ22 \cos 2 \theta \cos \frac{\theta}{2}=2 \cos 3 \theta \cos \frac{9 \theta}{2}2cos2θcos2θ=2cos3θcos29θ cos5θ2+cos3θ2=cos15θ2+cos3θ2cos5θ2=cos15θ215θ2=2nπ±5θ215θ2±5θ2=2nπ10θ=2nπ or 5θ=2nπθ=nπ5 or θ=2nπ5⇒θ=nπ5\begin{aligned} & \cos \frac{5 \theta}{2}+\cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2}+\cos \frac{3 \theta}{2} \\\\ & \cos \frac{5 \theta}{2}=\cos \frac{15 \theta}{2} \\\\ & \frac{15 \theta}{2}=2 n \pi \pm \frac{5 \theta}{2} \\\\ & \frac{15 \theta}{2} \pm \frac{5 \theta}{2}=2 n \pi \\\\ & 10 \theta=2 n \pi \quad \text { or } 5 \theta=2 n \pi \\\\ & \theta=\frac{n \pi}{5} \text { or } \theta=\frac{2 n \pi}{5} \\\\ & \Rightarrow \theta=\frac{n \pi}{5} \end{aligned}cos25θ+cos23θ=cos215θ+cos23θcos25θ=cos215θ215θ=2nπ±25θ215θ±25θ=2nπ10θ=2nπ or 5θ=2nπθ=5nπ or θ=52nπ⇒θ=5nπ θ=±π,±4π5,±3π5,±2π5,±π5m=5,n=5mn=25\begin{aligned} & \theta=\pm \pi, \pm \frac{4 \pi}{5}, \pm \frac{3 \pi}{5}, \pm \frac{2 \pi}{5}, \pm \frac{\pi}{5} \\\\ & m=5, \quad n=5 \\\\ & m n=25 \end{aligned}θ=±π,±54π,±53π,±52π,±5πm=5,n=5mn=25