Question

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

Answer 1

Разделим на cos²x
2tg²x+3tgx-2=0
tgx=a
2a²+3a-2=0
D=9+16=25
a1=(-3-5)/4=-2⇒tgx=-2⇒x=-arctg2+πn,n∈z
a2=(-3+5)/4=0,5⇒tgx=0,5⇒x=arctg0,5+πk,k∈z