$2\cos^{3}x + cos2x+4sinx-3=0$
$pt<=>2cos^3x+2cos^2x+4sinx-4=0$

$<=>cos^2x(cosx+1)+2(sinx-1)=0$

$<=>(1-sin^2x)(cosx+1)+2(sinx-1)=0$

$<=>(1+sinx)(1-siinx)(cosx+1)+2(sinx-1)$=0

$<=>(sinx-1)[2-(1+sinx)(1+cosx)]=0$

$<=>sinx=1$ 

hoặc $2-1-sinx-cosx-sinxcosx=0$

đặặt $t=sinx+cosx=>sinxcosx=\frac{1-t^2}{2}$

$xong$
biến đổi vòng vèo –  Nero 22-05-14 08:43 PM
Pt $\Leftrightarrow 2cosx(1-sin^2x)-2-2sin^2x+2sinx+2sinx=0$
$\Leftrightarrow 2cosx(1-sinx)(1+sinx)-2(1-sinx)+2sinx(1-sinx)=0$
$\Leftrightarrow 2(1-sinx)(cosx+sinx.cosx+sinx-1)=0$
Dạng cơ bản hết rồi!
H chi con cach vote cho nhau thoi. Kk. Hoi toi luc tra loi thi xem xong roi thoi :d –  Nero 23-05-14 09:55 AM
mi k duoc vote :)) ta vote cho mi –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 23-05-14 07:04 AM
Thấy đúng thì tích vào biểu tượng chữ V màu trắng bên dưới vote down nhé! Lần sau mình sẽ ss giúp đỡ. Tks bạn! –  Nero 22-05-14 08:51 PM

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