Предмет: Геометрия, автор: Dmitriy64

Решите уравнение
4sin^2x+9sinxcosx+9cos^2x=2

Ответы

Автор ответа: mukus13

$4sin^2x+9sinxcosx+9cos^2x=2(cos^2x+sin^2x)$
$4sin^2x+9sinxcosx+9cos^2x=2cos^2x+2sin^2x$
$2sin^2x+9sinxcosx+7cos^2x=0$
делим все на cos²x≠0
$2tg^2x+9tgx+7=0$
замена tgx= t
$2t^2+9t+7=0$
D=81-56=25
t1= - 1
t2= - 3.5
tgx= - 1 или tgx= - 3.5
$x= - \frac{ \pi }{4} + \pi k,$ k∈Z или x= - arctg 3.5+πn, n∈Z

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