<=>$cos2x+\sqrt{3}sin2x-3(sinx+\sqrt{3}cosx)+2=0$
<=>$\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x-3(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx)+1=0$
<=>$cos(2x-\frac{\pi}{3})-3cos(x-\frac{\pi}{6})+1=0$
đặt $cos(x-\frac{\pi}{6})=cost$
=>$cos2t-3cost+1=0$
<=>$2cos^2t-3cost=0$
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