1+tg2(x/2)>0
2·(1+tg2(x/2))+(1–tg2(x/2))=2tg(x/2)·(1+tg2(x/2)
2+2tg2(x/2)+1–tg2(x/2)=2tg(x/2)+2tg3(x/2)
2tg3(x/2)–tg2(x/2)+2tg(x/2)–3=0
2tg3(x/2)–2 – (tg2(x/2)–1)2=0
2·(tg(x/2)–1)·(tg2(x/2)+tg(x/2)+1) – (tg2(x/2)–1)2=0
(tg(x/2)–1)·(2tg2(x/2)+2tg(x/2)+2–tg(x/2)+1)=0
tg(x/2)=1
(x/2)=(π/4)+πn, n ∈ Z
x= (π/2)+2πn, n ∈ Z
или
2tg2(x/2)+2tg(x/2)+2–tg(x/2)+1=0
2tg2(x/2)+(tg(x/2)+3=0
D < 0
нет корней
О т в е т. (π/2)+2πn, n ∈ Z