Question

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

Answer 1

$\frac{\tan x}{\cos2x}+4\sin^2x=0 \\ \cos2x\neq0==\ \textgreater \ \ 2x\neq\frac\pi2+\pi n;\ \ n\in Z;\\ x\neq\frac\pi4+\frac{\pi n}2;\ \ n\in Z;\\ \frac{\tan x+4\sin^2x\cos2x}{\cos2x}=0;\\ \tan x+4\sin^2x\cos2x=0;\\ \frac{\sin x}{\cos x}+4\sin^2x\cos2x=0;\\ \sin x\left(\frac1{\cos x}+4\sin x\cos2x\right)=0;\\ \frac{\sin x}{\cos x}\left(1+4\sin x\cos x\cos2x\right)=0;\\ \tan x\left(1+2\sin2x\cos2x\right)=0;\\ \tan x\left(1+\sin4x\right)=0;\\ a) \tan x=0==\ \textgreater \ x=\pi n;\ n\in Z;\\ b)1+\sin4x=0;\\ \sin4x=-1;\\$
$4x=-\frac\pi4+2\pi n;\\ x=-\frac{\pi}{16}+\frac{\pi n}{2};\\$