2sin^2x - (2+√2) sinxcosx + √2cos^2x=0

1. Разделим обе части на cos^2x:

2sin^2x - (2 + √2) sinx * cosx + √2cos^2x = 0; 2tg^2x - (2 + √2) tgx + √2 = 0.

2. Решим уравнение относительно tgx:

D = (2 + √2) ^2 - 4 * 2 * √2 = 4 + 4√2 + 2 - 8√2 = 4 - 4√2 + 2 = (2 - √2) ^2; tgx = ((2 + √2) ± √ (2 - √2) ^2) / 2 = ((2 + √2) ± (2 - √2)) / 2;

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

x = arctg√2 + πk, k ∈ Z;

2) tgx = ((2 + √2) + (2 - √2)) / 2 = (2 + √2 + 2 - √2) / 2 = 4/2 = 2;

x = arctg2 + πk, k ∈ Z.

Ответ: arctg√2 + πk; arctg2 + πk, k ∈ Z.