Question

2sin^2x+3sinxcosx+cos^2x=0 срочно родные

Answer 1

2sin²x+3sinxcosx+cos²x=0  /cos²x≠0
2tg²x+3tgx+1=0
tgx=a
2a²+3a+1=0
D=9-8=1
a1=(-3-1)/4=-1⇒tgx=-1⇒x=-π/4+πn
a2=(-3+1)/4=-1/2⇒tgx=-1/2⇒x=-arctg1/2+πn