a)$PT\Leftrightarrow2.2.(\frac{\sqrt{3}}{2}sin x+\frac{1}{2}cos x)=2\sqrt{3}(\frac{\sqrt{3}}{2}sin 2x+\frac{1}{2}cos 2x)$$\Leftrightarrow 4sin(x-\frac{\pi }{6})=2\sqrt{3}cos(2x-\frac{\pi }{3})$Để ý $2x-\frac{\pi }{3}=2(x-\frac{\pi }{6})$Từ đó đưa về pt bậc 2 vs sin
a)$PT\Leftrightarrow2.2.(\frac{\sqrt{3}}{2}sin x
-\frac{1}{2}cos x)=2\sqrt{3}(\frac{\sqrt{3}}{2}sin 2x+\frac{1}{2}cos 2x)$$\Leftrightarrow 4sin(x-\frac{\pi }{6})=2\sqrt{3}cos(2x-\frac{\pi }{3})$Để ý $2x-\frac{\pi }{3}=2(x-\frac{\pi }{6})$Từ đó đưa về pt bậc 2 vs sin