ĐK: \(x\ne-\dfrac{\pi}{2}+k2\pi;x\ne\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)
\(\dfrac{cosx-sin2x}{2cos^2x-sinx-1}=\sqrt{3}\)
\(\Leftrightarrow\dfrac{cosx-sin2x}{cos2x-sinx}=\sqrt{3}\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}.cos2x-\sqrt{3}.sinx\)
\(\Leftrightarrow cosx+\sqrt{3}.sinx=\sqrt{3}.cos2x+sin2x\)
\(\Leftrightarrow\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}.sinx=\dfrac{\sqrt{3}}{2}.cos2x+\dfrac{1}{2}sin2x\)
\(\Leftrightarrow sin\dfrac{\pi}{6}.cosx+cos\dfrac{\pi}{6}.sinx=sin\dfrac{\pi}{3}.cos2x+cos\dfrac{\pi}{3}.sin2x\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=sin\left(2x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{6}=2x+\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{6}=\pi-2x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}-k2\pi\left(tm\right)\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\left(l\right)\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{6}-k2\pi\)
