$2sinx.(1+cos2x)+sin2x=1+2cosx$
Cách giải khác ^,^
$\Leftrightarrow 2\sin x(\sin^{2}x+\cos^{2}x+\cos^{2}x-\sin^{2}x)-2\cos x+(\sin 2x-1)=0$
$\Leftrightarrow 4\sin x\cos^{2}x-2\cos x+(\sin 2x-1)=0$
$\Leftrightarrow 2cosx(2\sin x\cos x-1)+(\sin 2x-1)=0$
$\Leftrightarrow (\sin 2x-1)(2\cos x+1)=0$
Bạn tự làm nốt nhé. Lời giải đúng thì vote cho mình nhé!
$2sinx(1+cos2x)+sin2x=1+2cosx$

$\Leftrightarrow 2sinx+2sinxcos2x+sin2x=1+2cosx$

$\Leftrightarrow 2sinx(cosx-sinx)(cosx+sinx)-2(cosx-sinx)-(1-sin2x)=0$

$\Leftrightarrow 2sinx(cosx-sinx)(cosx+sinx)-2(cosx-sinx)-(cosx-sinx)^2=0$

$\Leftrightarrow (cosx-sinx)[2sinx(sinx+cosx)-2-cosx+sinx]=0$

$\Leftrightarrow (cosx-sinx)(2sin^2x+2sinxcosx-2-cosx+sinx)=0$

$\Leftrightarrow (cosx-sinx)(2-2cos^2x+2sinxcosx-2-cosx+sinx)=0$

$\Leftrightarrow (cosx-sinx)[-2cosx(cosx-sinx)-(cosx-sinx)]=0$

$\Leftrightarrow (cosx-sinx)^2(-2cosx-1)=0$

Còn lại bạn giải bình thường là ra
Nếu đúng bạn nhấn V để chấp nhận, nhấn mũi tên để vote up cho đáp án của mình nha. Cảm ơn :) –  NguyễnTốngKhánhLinh 14-07-13 10:38 PM

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