ĐKXĐ: \(cosx\ne\pm\frac{1}{2}\)
\(4sinx+4\sqrt{3}cosx-2\sqrt{3}sin2x-3=2cos2x-1\)
\(\Leftrightarrow2sinx+2\sqrt{3}cosx-\sqrt{3}sin2x-cosx=1\)
\(\Leftrightarrow2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x\right)=\frac{1}{2}\)
\(\Leftrightarrow2sin\left(x+\frac{\pi}{3}\right)-sin\left(2x+\frac{\pi}{6}\right)=\frac{1}{2}\)
Đặt \(x+\frac{\pi}{3}=t\Rightarrow x=t-\frac{\pi}{3}\Rightarrow2x+\frac{\pi}{6}=2t-\frac{\pi}{2}\)
Pt trở thành:
\(2sint-sin\left(2t-\frac{\pi}{2}\right)=\frac{1}{2}\)
\(\Leftrightarrow2sint+cos2t=\frac{1}{2}\)
\(\Leftrightarrow2sint+1-2sin^2t=\frac{1}{2}\)
\(\Leftrightarrow...\)
