Question

$Sin ^{2}x+Sin 4x=-Cos ^{2}x$

Answer 1

$sin^2x+sin4x=-cos^2x$
ОДЗ: x∈R
$sin^2x+cos^2x+sin4x=0$
$1+sin4x=0$
$sin4x=-1$
$4x=- \frac{ \pi }{2} +2 \pi n;$ n∈Z (:4)
$x=- \frac{ \pi }{8} + \frac{ \pi n}{2} ;$ n∈Z