\(sin^3x+cos^3x=\dfrac{\sqrt{2}}{2}\) (*)
\(\Rightarrow\) \(\left(sinx+cosx\right)\left(sin^2x-sinx\cdot cosx+cos^2x\right)=\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow\) \(\left(sinx+cosx\right)\left(1-sinx\cdot cosx\right)=\dfrac{\sqrt{2}}{2}\) (1)
Đặt \(t=sinx+cosx\left(t\le\left|\sqrt{2}\right|\right)\)
\(\Rightarrow\)\(sinx\cdot cosx=\dfrac{t^2-1}{2}\)
Khi đó (1) thành: \(t\cdot\dfrac{1-t^2}{2}=\dfrac{\sqrt{2}}{2}\) \(\Rightarrow t=-\sqrt{2}\left(tm\right)\)
\(\Rightarrow sinx+cosx=-\sqrt{2}\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{2}\)
\(\Rightarrow sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Rightarrow x+\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\) \(\Rightarrow x=-\dfrac{3\pi}{4}+k2\pi\left(k\in Z\right)\)
