Sin²x + 6sinxcosx + 8cos²x=0 3tgx – 6ctgx + 7=0 - вопрос №34272346803670 от grrrrrrrrrr 17.11.2021 05:49

Question

Алгебра: Sin²x + 6sinxcosx + 8cos²x=0 3tgx – 6ctgx + 7=0

grrrrrrrrrr · Answer

1)sin²x+6sinxcosx+8cos²x=0 /cos²x≠0tg²x+6tgx+8=0tg²x=aa²+6a+8=0a1+a2=-6 U a1*a2=8a1=-4⇒tgx=-4⇒x=-arctg4+πna2=-2⇒tgx=-2⇒x=-arctg2+πn3tgx-6ctgx+7=03tgx-6/tgx+7=03tg²x+7tgx-6=0      tgx≠0tgx=a3a²+7a-6=0D=49+72=121a1=(-7-11)/6=-3⇒tgx=-3⇒x=-arctg3+πna2=(-7+11)/6=2/3⇒tgx=2/3⇒x=arctg2/3+πn...