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(A) Given equation: $\cos 2 	heta \cos \frac{	heta}{2} = \cos 3 	heta \cos \frac{9 	heta}{2}$.Multiply by $2$ on both sides: $2 \cos 2 	heta \cos

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) Given equation: $\cos 2 	heta \cos \frac{	heta}{2} = \cos 3 	heta \cos \frac{9 	heta}{2}$.Multiply by $2$ on both sides: $2 \cos 2 	heta \cos \frac{	heta}{2} = 2 \cos \frac{9 	heta}{2} \cos 3 	heta$.Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we get:$\cos \frac{5 	heta}{2} + \cos \frac{3 	heta}{2} = \cos \frac{15 	heta}{2} + \cos \frac{3 	heta}{2}$.$\cos \frac{15 	heta}{2} = \cos \frac{5 	heta}{2}$.General solution: $\frac{15 	heta}{2} = 2 k \pi \pm \frac{5 	heta}{2}$.Case $1$: $\frac{15 	heta}{2} - \frac{5 	heta}{2} = 2 k \pi$ $\Rightarrow 5 	heta = 2 k \pi$ $\Rightarrow 	heta = \frac{2 k \pi}{5}$.Case $2$: $\frac{15 	heta}{2} + \frac{5 	heta}{2} = 2 k \pi$ $\Rightarrow 10 	heta = 2 k \pi$ $\Rightarrow 	heta = \frac{k \pi}{5}$.Combining both,$	heta = \frac{k \pi}{5}$ for $k \in \mathbb{Z}$.In $[-\pi, \pi]$,$	heta \in \{-\pi, -\frac{4 \pi}{5}, -\frac{3 \pi}{5}, -\frac{2 \pi}{5}, -\frac{\pi}{5}, 0, \frac{\pi}{5}, \frac{2 \pi}{5}, \frac{3 \pi}{5}, \frac{4 \pi}{5}, \pi\}$.Positive values $(m)$: $\{\frac{\pi}{5}, \frac{2 \pi}{5}, \frac{3 \pi}{5}, \frac{4 \pi}{5}, \pi\}$,so $m = 5$.Negative values $(n)$: $\{-\pi, -\frac{4 \pi}{5}, -\frac{3 \pi}{5}, -\frac{2 \pi}{5}, -\frac{\pi}{5}\}$,so $n = 5$.Therefore,$mn = 5 	imes 5 = 25$.

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

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