ĐKXĐ: \(sinx\ne\frac{1}{2}\Rightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{6}+k2\pi\\x\ne\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow2\sqrt{3}sinx+2\sqrt{3}sinx.cosx-2cosx\left(1-cosx\right)-3=0\)
\(\Leftrightarrow2\sqrt{3}sinx-2cosx+\sqrt{3}sin2x+2cos^2x-1-2=0\)
\(\Leftrightarrow4\left(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx\right)+2\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x\right)-2=0\)
\(\Leftrightarrow2sin\left(x-\frac{\pi}{6}\right)+sin\left(2x+\frac{\pi}{6}\right)=1\)
Đặt \(x-\frac{\pi}{6}=a\)
\(2sina+sin\left(2a+\frac{\pi}{2}\right)=1\)
\(\Leftrightarrow2sina+cos2a=1\)
\(\Leftrightarrow2sina+1-2sin^2a=1\)
\(\Leftrightarrow sina\left(1-sina\right)=0\Rightarrow\left[{}\begin{matrix}sina=0\\sina=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{6}\right)=0\\sin\left(x-\frac{\pi}{6}\right)=1\end{matrix}\right.\) \(\Rightarrow...\)
Hơi dài, ko biết có đi đường vòng ở đoạn nào ko nữa