Question

Solve the following equation:
$1-\sin 2x=\cos x-\sin x$

Answer 1

$1-\sin 2x=\cos x-\sin x$

${(\cos x-\sin x)}^2=\cos x-\sin x$

So, either $\cos x-\sin x=1,\tan x=1,x=n\pi +\frac{\pi }{4}$

or, $\cos x-\sin x=1$

$\cos x-1=\sin x$

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

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

$x=2n\pi .$

or, $\tan \frac{x}{2}=-1$

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

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