The roots of the equation 1 - cos theta = sin | vedclass

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(B) Given equation: $1 - \cos 	heta = \sin 	heta \cdot \sin \frac{	heta}{2}$Using the identity $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $\

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$2k\pi, k \in I$

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(B) Given equation: $1 - \cos 	heta = \sin 	heta \cdot \sin \frac{	heta}{2}$Using the identity $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$:$2 \sin^2 \frac{	heta}{2} = (2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}) \cdot \sin \frac{	heta}{2}$$2 \sin^2 \frac{	heta}{2} = 2 \sin^2 \frac{	heta}{2} \cos \frac{	heta}{2}$$2 \sin^2 \frac{	heta}{2} (1 - \cos \frac{	heta}{2}) = 0$This implies $\sin^2 \frac{	heta}{2} = 0$ or $\cos \frac{	heta}{2} = 1$.If $\sin \frac{	heta}{2} = 0$,then $\frac{	heta}{2} = k\pi \implies 	heta = 2k\pi$.If $\cos \frac{	heta}{2} = 1$,then $\frac{	heta}{2} = 2k\pi \implies 	heta = 4k\pi$.Since $4k\pi$ is a subset of $2k\pi$ for $k \in I$,the general solution is $	heta = 2k\pi, k \in I$.

The roots of the equation $1 - \cos \theta = \sin \theta \cdot \sin \frac{\theta}{2}$ are:

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