ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne-1\\sinx\ne\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{cosx-sin2x}{cos2x-sinx}=\sqrt{3}\)
\(\Rightarrow cosx-sin2x=\sqrt{3}cos2x-\sqrt{3}sinx\)
\(\Leftrightarrow cosx+\sqrt{3}sinx=sin2x+\sqrt{3}cos2x\)
\(\Leftrightarrow\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=sin\left(2x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=x+\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\) (thỏa mãn ĐKXĐ)
