Question

Решите, пожалуйста, уравнение.
(tg (x)/cos (2x))+4sin^2 (x) =0

Answer 1

$frac{tan x}{cos2x}+4sin^2x=0 \ cos2xneq0== textgreater 2xneqfracpi2+pi n; nin Z;\ xneqfracpi4+frac{pi n}2; nin Z;\ frac{tan x+4sin^2xcos2x}{cos2x}=0;\ tan x+4sin^2xcos2x=0;\ frac{sin x}{cos x}+4sin^2xcos2x=0;\ sin xleft(frac1{cos x}+4sin xcos2xright)=0;\ frac{sin x}{cos x}left(1+4sin xcos xcos2xright)=0;\ tan xleft(1+2sin2xcos2xright)=0;\ tan xleft(1+sin4xright)=0;\ a) tan x=0== textgreater x=pi n; nin Z;\ b)1+sin4x=0;\ sin4x=-1;$
$4x=-fracpi4+2pi n;\ x=-frac{pi}{16}+frac{pi n}{2};$