4(sin4x+cos4x)+sin4x−2=0
<=>4(sin4x+cos4x)+2sin2xcos2x−2=0
<=>4(sin4x+cos4x)+4sinxcosx(cos2x−sin2x)−2=0(∗)
TH1:cosx=0<=>x=π2+kπ(k∈Z)=>sinx=1 hoac sinx=−1
thay vao (∗) ta duoc 2=0=>x=π2+kπ(k∈Z) k la nghiem cua pt
TH2cosx≠0 chia hai ve cua (∗) cho cos4x duoc
4(tan4x+1)+4tanx−4tan3x−2(1+tan2x)2=0
<=>4tan4x+4+4tanx−4tan3x−2tan2x−4tanx−2=0
<=>2tan4x−4tan3x+2=0