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

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

Question

(b) We have, $1 - \cos 	heta = \sin 	heta \,.\,\sin \frac{	heta }{2}$

==> $2{\sin ^2}\frac{	heta }{2} = 2\sin \frac{	heta }{2}\,.\,\cos \fr

Vedclass · Accepted Answer

$2k\pi ,k \in I$

Vedclass · Answer

(b) We have, $1 - \cos 	heta = \sin 	heta \,.\,\sin \frac{	heta }{2}$

==> $2{\sin ^2}\frac{	heta }{2} = 2\sin \frac{	heta }{2}\,.\,\cos \frac{	heta }{2}\,.\,\sin \frac{	heta }{2}$

==> $2{\sin ^2}\frac{	heta }{2}\,\left[ {1 - \cos \frac{	heta }{2}} ight] = 0$

==> $\sin \frac{	heta }{2} = 0$ or $2{\sin ^2}\frac{	heta }{4} = 0$

==> $\sin \frac{	heta }{2} = 0$ or $\sin \frac{	heta }{4} = 0$

==> $\frac{	heta }{2} = k\pi $ or $\frac{	heta }{4} = k\pi $.

Hence, $	heta = 2k\pi $ or $	heta = 4k\pi $, $k \in I$.

The roots of the equation $1 - \cos \theta = \sin \theta .\sin \frac{\theta }{2}$ is

Similar Questions

If $\frac{{\tan 3\theta - 1}}{{\tan 3\theta + 1}} = \sqrt 3 $, then the general value of $\theta $ is

If $\theta $ and $\phi $ are acute satisfying $\sin \theta = \frac{1}{2},$ $\cos \phi = \frac{1}{3},$ then $\theta + \phi \in $

If $\cos 2\theta + 3\cos \theta = 0$, then the general value of $\theta $ is

The equation $5x^2+12x + 13 = 0$ and $ax^2+bx + c = 0$ have a common root, where $a,b,c$ are the sides of $\Delta ABC$,then find $\angle C$ ? .....$^o$

No. of solution of equation $sin^{65}x\, -\, cos^{65}x =\, -1$ is, if $x \in (-\pi , \pi )$