Question

Solve the following equations.
$\frac{1-cos2x}{sin2x}=0.$

Answer 1

$\frac{1-\cos 2x}{\sin 2x}=0=>1-\cos 2x=0;\sin 2x\ne 0=>\cos 2x=1;2x\ne m\pi =>2x=2n\pi ;x\ne \frac{m\pi }{2}=>x=n\pi$

$m,n$ belongs to integers.