tìm $m$ để pt $(cosx+1)(cos2x-mcosx)=msin^{2}x$ có nghiệm thuộc đoạn $[0; \frac{2\pi}{3}]$
pt$\Leftrightarrow (cosx+1)(cos2x-mcosx)-m(1-cos^2x)=0$
$\Leftrightarrow (cosx+1)(cos2x-mcosx-m(1-cosx))=0$
$\Leftrightarrow (cosx+1)(cos2x-m)=0$
pt da cho  co nghiem $x\in[0;\frac{2\pi}{3}]$ khi $cos2x=m $ co nghiem $x\in [0;2\pi/3]$
Ta co:$0\leq x\leq 2\pi/3\Leftrightarrow 0\leq 2x\leq 4\pi/3$
khi do $-1\leq cos2x\leq 1$hay $-1\leq m\leq 1$
Vay $m\in[-1;1]$
gio k ke duoc cai bang nen phai xai` cach nay`,hi`^^ –  Gió! 04-12-13 08:08 AM
ko có troll đây nhé chỉ là sao gios ko dg` bảng biến thiên cho nhanh –  doremon1191997 04-12-13 07:56 AM
k biet co dung hay k nua^^:P –  Gió! 03-12-13 07:57 PM

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