Найдите sin 2x если 0

Tg (x+π/4) - 2tgx=2

(tgx+tgπ/4) / (1-tgx*tgπ/4) - 2tgx=2

(tgx+1) / (1-tgx) - 2tgx=2

(tgx+1) / (1-tgx) - 2 (tgx+1) = 0

(tgx+1) * (1-2+2tgx) / (1-tgx) = 0

(tgx+1) (2tgx-1) = 0, tgx≠1

tgx+1=0⇒tgx=-1⇒x=-π/4+πn не удов усл

2tgx-1=0⇒tgx=1/2

cos²x=1: (1+tg²x) = 1: (1+1/4) = 1:5/4=4/5

cosx=2/√5

sinx=√ (1-cos²x) = √ (1-4/5) = √ (1/5) = 1/√5

sin2x=2sinxcosx=2*1/√5*2/√5=4/5