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Question

$cos2x-tan^2x=\frac{cos^2x-cos^3x-1}{cos^2x}$

$sinxcos4x+2sin^2x=1-4sin^2(\frac{\pi}4-\frac{x}4)$

mk lam duoc roi va co giao chua roi. nhung de mn lam.

score 2 · Answer 1

Câu 1

$cos2x-tan^2x=\frac{-(1-cos^2x)-cos^3x}{cos^2x}$

$<=>cos2x-tan^2x=-\frac{sin^2x}{cos^2x}-\frac{cos^3x}{cos^2x}$

$<=>cos2x + cosx=0$

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

Dạng cơ bản rồi

cuonsachnho1 · Answer 2 · 2/22/2014 6:13:13 PM

Câu 1:

<=>$\cos ^2x$-$\sin ^2x$-$\sin^2 x$/$\cos^2 x$=$\cos x$-1/$\cos x$-$\cos ^2x$

<=>$\cos^2 x$-1/$\cos^2 x$-$\sin^2 x$+1=$\cos x$-1/$\cos x$-$\cos ^2x$

<=>$($cos x$-1/$cos x$)^{2}$-2+2$\cos^2 x$=$\cos x$-1/$\cos x$

<=>($\cos^2 x$-1)[($\cos^2 x$-1)/$\cos^2 x$ + 2 -1/$\cos x$]=0.

Trả lời 22-02-14 06:13 PM

cuonsachnho1
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sao mà k hieu?b k hieu chỗ nào ? – cuonsachnho1 22-02-14 08:58 PM

k hieu bai cua ban. – ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 22-02-14 07:08 PM

bn tự giải not nhá . – cuonsachnho1 22-02-14 06:14 PM

score 0 · Answer 3

1)PT$\Leftrightarrow \cos 2x=\frac{\cos^2 x-\cos^3 x-1}{\cos^2 x}+\frac{\sin^2 x}{\cos^2 x}$

$ \Leftrightarrow \cos 2x=\frac{\cos x(\cos ^2x-cos^3x-1)+sin^2x}{cos^2x}$

$\Leftrightarrow cos 2x=\frac{cos^3x-cos^4x-1+sin^2x}{cos^2x}$

$\Leftrightarrow cos 2x=\frac{cos^3x-cos^4x-1+1-cos^2x}{cos^2x}$

$\Leftrightarrow cos 2x=\frac{cos^3x-cos^4x-cos^2x}{cos^2x}$

$\Leftrightarrow cos 2x=cosx-cos^2x-1$

$\Leftrightarrow cos^2x-sin^2x=cosx-cos^2x-1$

$\Leftrightarrow cos^2x-cosx+2=0$

Đến đây để rồi bn tự giải nhé! Good luck!

sai roi nhe. dong thu 3. voi lai co giao mk chep nham de bai len bang len cau 1 sai de bai – ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 22-02-14 07:07 PM

lam` sai tum` lum het' kia` >"< ( o cho dau tuong duong thu 1 va 3 ay) – Gió! 22-02-14 06:51 PM

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