2cos^2x-3sinx cosx+sin^2x=0

2cos^2 (x) - 3sin (x) cos (x) + sin^2 (x) = 0;

Делим на cos^2 (x) ≠ 0;

2 - 3tg (x) + tg^2 (x) = 0;

Делаем замену переменной:

tg (x) = t;

t^2 - 3t + 2 = 0;

D = 1;

t1 = (3 + 1) / 2 = 2;

t2 = (3 - 1) / 2 = 1;

tg (x) = 2 = => x = arctg (2) + пn, n∈Z.

tg (x) = 1 = => x = п/4 + пk, k∈Z.

Ответ: x = arctg (2) + пn, n∈Z или x = п/4 + пk, k∈Z.