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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $\

Vedclass · Accepted Answer

$2k\pi, k \in I$

Vedclass · Answer

(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:

Similar Questions

The number of solutions of $\sqrt{\tan \theta} = 2 \sin \theta$ for $\theta \in [0, 2\pi]$ is equal to:

If $\cos x = |\sin x|$,then the general solution is

What is the number of roots of the quadratic equation $8\sec^2\theta - 6\sec\theta + 1 = 0$?

The general solution of $\sin x + \cos x = 1$ is

If the general solution set of $\sin x + 3 \sin 3x + \sin 5x = 0$ is $S$,then $\{\sin \alpha \mid \alpha \in S\} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module