|
|
mãi mới ra gê qá X_X
$pt <=> Sin3x+Cos3x-2Sinx= 2\sqrt{2}(1-4\sin ^2x$)
$<-> Sin3x-Sinx-Sinx+Cos3x= 2\sqrt{2}(1-4(1-Cos^2x))
<-> 2Cos2xSinx-Sinx +Cos3x= 2\sqrt{2}(4Cos^2x-3)$
$ <->Sinx(2Cos2x-1)+Cos3x= 2\sqrt{2}(4Cos^2x-3)$
<-> $Sinx\left[ {2(2Cos^2x-1)-1})+Cos3x= 2\sqrt{2}(4Cos^2x-3)\right.$
$<->(Sinx-2\sqrt{2})(4Cos^2x-3)+Cosx3=0$
$<->(Sinx-2\sqrt{2})(4Cos^2x-3)+Cos(4Cos^2x-3)=0$
$<->(4Cos^2x-3)(Sinx+Cosx -2\sqrt{2})=0$
|