{–tgxtgy=1/3;
{y=π–x,
{–tgx·tg(π–x)=1/3;
{y=π–x,
{–tgx·(–tgx)=1/3;
{y=π–x,
{tg2 x=1/3;
{y=π–x,
{tgx= ± (√3)/3;
1) tgx=–(√3)/3,
x=arctg(–(√3)/3)+πk, k ∈ Z,
x1=–(π/6)+πk, k ∈ Z,
y1=π+(π/6)–πk=(7π)/6–πk, k ∈ Z;
2) tgx=(√3)/3,
x=arctg((√3)/3)+πn, n ∈ Z,
x2=(π/6)+πn, n ∈ Z,
y2=π–(π/6)–πn=(5π)/6–πn, n ∈ Z.
Ответ: (–(π/6)+πk;(7π)/6–πk), k ∈ Z;
((π/6)+πn; (5π)/6–πn), n ∈ Z.