$4(sin^4x+cos^4x)+sin4x-2=0$
$<=>4(sin^4x+cos^4x)+2sin2xcos2x-2=0$
$<=>4(sin^4x+cos^4x)+4sinxcosx(cos^2x-sin^2x)-2=0(*)$
$TH1:cosx=0<=>x=\frac{\pi}{2}+k\pi(k \in Z)=>sinx=1$ hoac $sinx=-1$
thay vao $(*)$ ta duoc $2=0=>x=\frac{\pi}{2}+k\pi(k\in Z)$ k la nghiem cua pt
$TH2cosx\neq 0$ chia hai ve cua $(*)$ cho $cos^4x$ duoc
$4(tan^4x+1)+4tanx-4tan^3x-2(1+tan^2x)^2=0$
$<=>4tan^4x+4+4tanx-4tan^3x-2tan^2x-4tanx-2=0$
$<=>2tan^4x-4tan^3x+2=0$