Question

Solve the following equation:
$\tan \frac{x}{2}=1-\cos x$

Answer 1

$\tan \frac{x}{2}=1-\frac{(1-{\tan }^2\frac{x}{2})}{(1+{\tan }^2\frac{x}{2})}$

Let $\tan \frac{x}{2}=z.$

$z=1-\frac{(1-z^2)}{(1+z^2)}$

$z(1+z^2)=1+z^2-1+z^2$

$z+z^3={2z}^2$

$z^2+1-2z=0$

${(z-1)}^2=0$

$z=1$

$\tan \frac{x}{2}=1$

$\frac{x}{2}=n\pi +\frac{\pi }{4}$

$x=2n\pi +\frac{\pi }{2}$

and, $\tan \frac{x}{2}=0$

$\frac{x}{2}=n\pi$

$x=2n\pi .$