Question

3 sin^2 x = 2 sin x cos x + cos^2 x

Answer 1

3 sin^2 x - 2 sin x cos x - cos^2 x = 0;    /:cos^2 x ≠ 0;
3 tg^2 x - 2 tg x - 1 = 0;
 tg x = a ; 
3 a^2 - 2 a - 1 = 0;
D = 4 + 12 =16= 4^2;
 a1 = (2+4) / 6 = 1; ⇒ tg x = 1;  x = pi/4 + pi*k; k-Z;
a2= (2-4) /6= - 1/3 ; ⇒ tg x = - 1/3;  x = - arctg(1/3) + pi*k; k-Z