Question

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