Question

sin^2x+2sin2x+3cos^2x=0

Answer 1

Разделим на cos²x
tg²x+4tgx+3=0
tgx=a
a²+4a+3=0
a1+a2=-4 U a1*a2=3
a1=-3⇒tgx=-3⇒x=-arctg3+πn,n∈z
a2=-1⇒tgx=-1⇒x=-π/4+πk,k∈z