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JEE Main 2020
Trigonometric Equations
Trigonometric Equations
Hard

Question

If m and n respectively are the numbers of positive and negative values of θ\theta in the interval [π,π][-\pi,\pi] that satisfy the equation cos2θcosθ2=cos3θcos9θ2\cos 2\theta \cos {\theta \over 2} = \cos 3\theta \cos {{9\theta } \over 2}, then mn is equal to ____________.

Answer: 2

Solution

2cos2θcosθ2=2cos3θcos9θ22 \cos 2 \theta \cos \frac{\theta}{2}=2 \cos 3 \theta \cos \frac{9 \theta}{2} cos5θ2+cos3θ2=cos15θ2+cos3θ2cos5θ2=cos15θ215θ2=2nπ±5θ215θ2±5θ2=2nπ10θ=2nπ or 5θ=2nπθ=nπ5 or θ=2nπ5θ=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} 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}

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