Giai phương trình

                       $\sqrt{3}(\sin4 x-\cos 2x)=2\cos \left ( x+\frac{\Pi }{4} \right )\sin \left (3 x-\frac{3\Pi }{4} \right )$
$PT<=>\sqrt{3}(sin4x-cos2x)=sin(4x-\frac{\pi}{2})-sin(-2x+\pi)$

$<=>\sqrt3(sin4x-cos2x)=cos4x-sin2x$

$<=>\sqrt3sin4x-cos4x+sin2x-\sqrt3cos2x=0$

$<=>\frac{\sqrt3}{2}sin4x-\frac{1}{2}cos4x+\frac{1}{2}sin2x-\frac{\sqrt3}{2}cos2x=0$

$<=>sin(4x-\frac{\pi}{6})+sin(2x-\frac{\pi}{3})=0$

$<=>2sin(3x-\frac{\pi}4)cos(x+\frac{\pi}{12})=0$

$XONG$
tks ban. ban cung vay. –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 29-01-14 04:33 PM
Cám ơn bạn nhé! Chúc bạn và gia đình có 1 cái tết đầm ấm , hạnh phúc và 1 năm mới thịnh vượng như ý nhé! –  leejongsukleejongsuk 29-01-14 03:41 PM

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