Решите уравнение - sin ) ^2 - 1

$cos(x)=[cos( \frac{x}{2} )-sin( \frac{x}{2} )]^2-1$
$cos(x)=[sin( \frac{\pi}{2}+ \frac{x}{2} )-sin( \frac{x}{2} )]^2-1$
$cos(x)=[2sin( \frac{ \frac{\pi}{2} + \frac{x}{2}- \frac{x}{2}}{2} )cos(\frac{ \frac{\pi}{2} + \frac{x}{2}+ \frac{x}{2}}{2})]^2-1;$
$cos(x)=[2cos(\frac{\pi}{4} + \frac{x}{2})sin( \frac{\pi}{4} )]^2-1;$
$cos(x)+1=2cos^2(\frac{\pi}{4} + \frac{x}{2});$
$cos(x)+1=2*\frac{1+cos(\frac{\pi}{2} + x)}{2}$
$cos(x)+1=1+cos(\frac{\pi}{2} + x)$
$cos(x)+cos(\frac{\pi}{2} + x)=0;$
$2cos( \frac{\frac{\pi}{2}+ x-x}{2} )*cos( \frac{\frac{\pi}{2}+ x+x}{2} )=0$
$cos( \frac{\pi}{4}+ x )=0$
$\frac{\pi}{4}+ x = \frac{\pi}{2}+\pi n,n\in Z$
$x = \frac{\pi}{4}+\pi n,n\in Z$

Ответ: $\frac{\pi}{4}+\pi n,n\in Z$