Giải PT: $2cos4x - (\sqrt{3}-2)cos2x=sin2x+\sqrt{3}, x\in[0;\pi]$
$2cos4x - (\sqrt{3}-2)cos2x=sin2x+\sqrt{3}, x\in[0;\pi]$
Started By thanhthanhtoan, 26-07-2014 - 20:59
#1
Posted 26-07-2014 - 20:59
#2
Posted 26-07-2014 - 21:36
ta có Pt<=> $cos4x +Cos2x= \frac{\sqrt{3}}{2} Cos2x +\frac{1}{2} Sin 2x +\frac{\sqrt{3}}{2} = Cos \frac{\pi}{3}.Sin 2x + Sin \frac{\pi}{3}.Cos 2x + Sin \frac{\pi}{3}= Sin (2x+\frac{\pi}{3})+Sin \frac{\pi}{3} = 2Sin(x+\frac{\pi}{3}).Cos x $
do đó $2Cos3x.Cosx=2Sin(x+\frac{\pi}{3}).Cos x $
đến đây chắc ok giải pt Cosx=0 hoặc $Cos 3x =Sin(x+\frac{\pi}{3})$
Giải PT: $2cos4x - (\sqrt{3}-2)cos2x=sin2x+\sqrt{3}, x\in[0;\pi]$
ta có Pt<=> $cos4x +Cos2x= \frac{\sqrt{3}}{2} Cos2x +\frac{1}{2} Sin 2x +\frac{\sqrt{3}}{2} = Cos \frac{\pi}{3}.Sin 2x + Sin \frac{\pi}{3}.Cos 2x + Sin \frac{\pi}{3}= Sin (2x+\frac{\pi}{3})+Sin \frac{\pi}{3} = 2Sin(x+\frac{\pi}{3}).Cos x $
do đó $2Cos3x.Cosx=2Sin(x+\frac{\pi}{3}).Cos x $
đến đây chắc ok giải pt Cosx=0 hoặc $Cos 3x =Sin(x+\frac{\pi}{3})$
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