Question

срочно помогите 1-sinxcosx=sin^2x

Answer 1

sin²x+cos²x-sinxcosx-sin²x=0
cos²x-sinxcosx=0
cosx(cosx-sinx)=0
cosx=0⇒x=π/2+πn,n∈z
cosx-sinx=0/cosx
1-tgx=0
tgx=1
x=π/4+πk,k∈z