1-cos(x)=sin(x)*sin(x/2)

2sin^2x/2=2sin^2x/2cosx/2 sin^2x/2(1-cosx/2)=0 cosx/2=1 x/2=2пk x=4Пk sinx/2=0 x/2=Пk x=2Пk   x=2Пk  

1-(cos²x/2-sin²x/2)= 2·sinx/2·cosx/2·sinx/2  cos²x/2 + sin²x/2 - cos²x/2 + sin²x/2 = 2sin²x/2·cosx/2 2sin²x/2 - 2sin²x/2·cosx/2 = 0 2sin²x/2·( 1 - cosx/2) =0 sinx/2 = 0       или     1 - cosx/2 = 0 x/2 = πn, n∈Z,            cosx/2 = 1 х =  2πn, n∈Z,            x/2 = 2πk, k∈Z                                     х=4πk, k∈Z. Объединяя корни, получим, х =  2πn, n∈Z. Ответ:  2πn, n∈Z