cosx-sinx=1 помогите решить)

Cosx-sinx=0

(cos²x/2) - (sin²x/2) - 2 (sinx/2) (cosx/2) = 0 / - cos²x/2≠0

tg² (x/2) + 2tg (x/2) - 1=0

tgx/2=a

a2+2a-1=0

D=4+4=8

a1 = (-2-2√2) = 2=-1-√2⇒tgx/2=-1-√2⇒x/2=-arctg (1+√2) + πn⇒x=-2arctg (1+√2) + 2πn

a2 = (-2+2√2) = 2=-1=√2⇒tgx/2=-1+√2⇒x/2=arctg (-1+√2) + πn⇒x=2arctg (-1+√2) + 2πn