$\sqrt{cos2x}+\sqrt{1-sin2x}=2\sqrt{sinx-cosx}$
Edited by TianaLoveEveryone, 13-08-2015 - 22:56.
$\sqrt{cos2x}+\sqrt{1-sin2x}=2\sqrt{sinx-cosx}$
Edited by TianaLoveEveryone, 13-08-2015 - 22:56.
$\sqrt{cos2x}+\sqrt{1-sin2x}=2\sqrt{sinx-cosx}$
$\sqrt{(cosx-sinx)(cosx+sinx)}+\sqrt{(cosx-sinx)^2}=2\sqrt{sinx-cosx}\Leftrightarrow \sqrt{cosx-sinx}(\sqrt{cosx+sinx}+\sqrt{cosx-sinx}-2)=0$
TH1:sinx=cosx
TH2$\sqrt{cosx+sinx}+\sqrt{cosx-sinx}=2\Leftrightarrow 2cosx+2\sqrt{cos^2x-sin^2x}=2\Leftrightarrow 2cos^2x-1=(1-cos)^2$
Đến đây thì bạn tự làm tiếp nhé.
0 members, 1 guests, 0 anonymous users