ta có : \(sin^3x.cosx-cos^3x.sinx=\dfrac{\sqrt{2}}{8}\)
\(\Leftrightarrow sinx.cosx\left(sin^2x-cos^2x\right)=\dfrac{\sqrt{2}}{8}\)
\(\Leftrightarrow\dfrac{-1}{2}sin2x.cos2x=\dfrac{\sqrt{2}}{8}\Leftrightarrow sin2x.cos2x=-\dfrac{1}{2\sqrt{2}}\)
\(\Leftrightarrow\dfrac{1}{2}sin4x=\dfrac{-1}{2\sqrt{2}}\Leftrightarrow sin4x=\dfrac{-\sqrt{2}}{2}=sin\left(\dfrac{-\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{-\pi}{4}+k2\pi\\4x=\pi+\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{16}+\dfrac{k\pi}{2}\\x=\dfrac{5\pi}{16}+\dfrac{k\pi}{2}\end{matrix}\right.\) \(\left(k\in Z\right)\)
vậy phương trình có 2 hệ nghiệm là \(x=\dfrac{-\pi}{16}+\dfrac{k\pi}{2}\) và \(x=\dfrac{5\pi}{16}+\dfrac{k\pi}{2}\) \(\left(k\in Z\right)\)
