$sin^2(\frac{x}{2}-\frac{\Pi }{4})+cos(x-\frac{\Pi }{6})=\frac{1}{8}(sin3x+sinx)(1+tan^2x)$
Bạn ơi, bạn để dấu $ ở hai đầu thì mới nhập công thức được nha :) –  NguyễnTốngKhánhLinh 20-06-13 09:00 AM
ĐK $cosx\neq 0$
Pt $\Leftrightarrow \frac{1-\cos (x-\pi/2)}{2}+\sqrt3/2.\cos x+1/2.\sin x=\frac{2\cos x\sin2x}{8\cos ^2x}$
$\Leftrightarrow 1-\sin x+\sqrt3\cos x+\sin x=\frac{4\sin x.\cos^2 x}{4\cos^2 x}$
$\Leftrightarrow 1+\sqrt3\cos x-\sin x=0$
$\Leftrightarrow 1/2\sin x-\sqrt3/2.\cos x=1/2 $
$\Leftrightarrow \sin (x-\pi/3)=1/2$
$\Leftrightarrow x=\pi/2+2k\pi$
hoặc $x=7\pi/6=2k\pi$
mình đã sửa lại nhưng khong thấy chỗ bạn nóimong bạn chỉ giúp !! –  Вő˚CĦĬĈŊ♂ 23-06-13 07:37 PM
bạn ơi đáp án này của bạn bị sai mất rồi, mình làm lại thấy chỗ sin3x sinx=2sin2xcosx chứ bạn –  tikki_likki 23-06-13 05:39 PM
dung rroi kich chu V ma chap nhan dap an cua nguoi ta –  ♂+♀→♥ 21-06-13 06:41 AM
ban oi dung roi thi kick vao chu V de xac nhan dap an di ! –  Вő˚CĦĬĈŊ♂ 20-06-13 08:08 PM
cảm ơn bạn nhiều nhé! –  tikki_likki 20-06-13 04:58 PM

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