Question

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

• A. $k\pi ,K\in 1$
• B. $2k\pi ,K\in 1$
• C. $\frac{k\pi }{2},K\in 1$
• D. none of these

Answer 1

The correct option is B $2k\pi ,K\in 1$

$1-\cos \theta =\sin \theta \sin \frac{\theta }{2}$

$\Rightarrow 1-1+2{\sin }^2\frac{\theta }{2}=\sin \theta \sin \frac{\theta }{2}$

$\Rightarrow 2{\sin }^2\frac{\theta }{2}-\sin \theta \sin \frac{\theta }{2}=0$

$\Rightarrow 2{\sin }^2\frac{\theta }{2}-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\sin \frac{\theta }{2}=0$

$\Rightarrow 2{\sin }^2\frac{\theta }{2}-2{\sin }^2\frac{\theta }{2}\cos \frac{\theta }{2}=0$

$\Rightarrow 2{\sin }^2\frac{\theta }{2}(1-\cos \frac{\theta }{2})=0$

$\Rightarrow 2{\sin }^2\frac{\theta }{2}=0$ or $1-\cos \frac{\theta }{2}=0$

$\Rightarrow {\sin }^2\frac{\theta }{2}=0$ or $\cos \frac{\theta }{2}=1$

$\Rightarrow \frac{\theta }{2}=k\pi$ or $\frac{\theta }{2}=2k\pi$

$\Rightarrow \theta =2k\pi$ or $\theta =4k\pi$ for $k\in Z$

$\therefore k=2\pi$ for $k\in Z$ is the solution of the equation