1-2sin x cos x + sin x + cos x = 0

sinx+cosx=t  (sinx+cosx)=t² sin²x+2sinxcosx+cos²x=1+2sinxcosx 2sinxcosx=t²-1 1-t²+1+t=0 t-t²=0 t(1-t)=0 t=0⇒sinx+cosx=0/cosx≠0 tgx+1=0⇒tgx=-1⇒x=-π/4+πn t=1⇒sinx+cosx=1 2sinx/2cosx/2+cos²x/2-sin²x/2-sin²x/2-cos²x/2=0 2sinx/2cosx/2-2sin²x/2=0 2sinx/2(cosx/2-sinx/2)=0 sinx/2=0⇒x/2=πn⇒x=2πn cosx/2-sinx/2=0/cosx/2≠0 1-tgx/4=0⇒tgx/4=1⇒x/4=π/4+πn⇒x=π+4πn Ответ x=-π/4+πn; x=2πn; x=π+4πn