The number of solutions of the equation cos 2 | vedclass

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(A) Given equation: $\cos 2 	heta \cos \frac{	heta}{2} + \cos \frac{5 	heta}{2} = 2 \cos^3 \frac{5 	heta}{2}$ Using the identity $2 \cos^3 A - \co

Vedclass · Accepted Answer

$7$

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(A) Given equation: $\cos 2 	heta \cos \frac{	heta}{2} + \cos \frac{5 	heta}{2} = 2 \cos^3 \frac{5 	heta}{2}$ Using the identity $2 \cos^3 A - \cos A = \cos 3A$,we rewrite the equation as: $\cos 2 	heta \cos \frac{	heta}{2} = 2 \cos^3 \frac{5 	heta}{2} - \cos \frac{5 	heta}{2} = \cos \left(3 	imes \frac{5 	heta}{2}ight) = \cos \frac{15 	heta}{2}$ Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,the $LHS$ becomes: $\frac{1}{2} \left( \cos \frac{5 	heta}{2} + \cos \frac{3 	heta}{2} ight) = \cos \frac{15 	heta}{2}$ This approach is complex; let us use the identity $\cos 3A = 4 \cos^3 A - 3 \cos A$. Actually,the equation simplifies to $\cos \frac{3 	heta}{2} = \cos \frac{15 	heta}{2}$. This implies $\frac{15 	heta}{2} = 2n \pi \pm \frac{3 	heta}{2}$. Case $1$: $\frac{15 	heta}{2} = 2n \pi + \frac{3 	heta}{2} \implies 6 	heta = 2n \pi \implies 	heta = \frac{n \pi}{3}$. For $	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$	heta \in \{-\frac{\pi}{3}, 0, \frac{\pi}{3}\}$. Case $2$: $\frac{15 	heta}{2} = 2n \pi - \frac{3 	heta}{2} \implies 9 	heta = 2n \pi \implies 	heta = \frac{2n \pi}{9}$. For $	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$	heta \in \{-\frac{4 \pi}{9}, -\frac{2 \pi}{9}, 0, \frac{2 \pi}{9}, \frac{4 \pi}{9}\}$. Combining both sets and removing duplicates $(0)$: $	heta \in \{-\frac{4 \pi}{9}, -\frac{\pi}{3}, -\frac{2 \pi}{9}, 0, \frac{2 \pi}{9}, \frac{\pi}{3}, \frac{4 \pi}{9}\}$. Total number of solutions is $7$.

The number of solutions of the equation $\cos 2 \theta \cos \frac{\theta}{2} + \cos \frac{5 \theta}{2} = 2 \cos^3 \frac{5 \theta}{2}$ in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is:

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