k

 $cos3x.cox^3x – sin3x.sin^3 x $= $ \frac{2+3\sqrt{2}}{8}$
chứng minh hay giải đây??? –  Confusion 25-04-16 01:01 PM
$cos3x = 4cos^3x-3cosx $
$sin3x = 3sinx -4sin^3x $
Xét vế trái
  $vt = cos^3x.(4cos^3x-3cosx) - sin^3x.(3sinx -4sin^3x) $
$=> vt = 4(cos^6x + sin^6x) - 3.(sin^4x +cos^4x) $
$=> vt =4.(sin^2x +cos^2x).(sin^4x -sin^2x.cos^2x + cos^4x) - 3.(sin^2x+cos^2x)^2 +6sin^2x.cos^2x $
$=> vt = 4.[(sin^2x+cos^2x)^2 - 3sin^2x.cos^2x] - 3 + 6sin^2x .cos^2x $
$=> vt = 1- 12sin^2x.cos^2x + 6sin^2x.cos^2x $
$=> vt = 1- 3/2.sin^2(2x) $
$=> vt = 1- 3/4.(1-cos4x) $
$=> 1- 3/4. + 3/4.cos4x = (2 + 3can2) / 8 $
$=> 2 + 6cos4x = 2+3\sqrt{2}$ 
$=> 6cos4x = 3\sqrt{2} $
$=> cos4x = 1\sqrt{2} $
$=> 4x= ± \pi 4 + k2pi $
$=> x= ± \pi 16 + k \pi 2$

quá khen –  tasfuskau 26-04-16 04:14 PM
ồ ôi giỏi kinh phết –  Đức Anh 25-04-16 10:36 PM

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