sin^3(x)+ cos^3(x)= 2sin^5(x)+2cos^5(x)
<=> sin^3(x)-2sin^5(x )+ cos^3(x)-2cos^5(x)=0
<=> sin^3(x) (1-2sin^2(x)) +cos^3(x) (1-2cos^2(x))=0
<=> sin^3(x) (1-2+2cos^2(x)) -cos^3(x) (2cos^2(x) -1 ) (chỗ này đặt dấu trừ ra ngoài) =0
<=> sin^3(x) (2cos^2(x)-1) -cos^3(x) (2cos(x)-1 )=0
<=> cos2x(sin^3(x) -cos^3(x))=0
=> | cos2x=0 <=> x=pi/4 +kpi/2
| sin^3(x)-cos^3(x)=0 <=> (sinx-cosx)(1+sinxcosx) =0 Đặt t= sinx-cosx => sinxcosx= - (t^2-1)/2
Giải t rồi triển oke