\(1+sin^3x+cos^3x=3sinx.cosx\)
\(\Leftrightarrow1+\left(sinx+cosx\right)\left(1-sinx.cosx\right)=3sinx.cosx\)
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
Pt trở thành:
\(1+t\left(1-\frac{t^2-1}{2}\right)=\frac{3}{2}\left(t^2-1\right)\)
\(\Leftrightarrow2+t\left(3-t^2\right)=3t^2-3\)
\(\Leftrightarrow t^3+3t^2-3t-5=0\)
\(\Leftrightarrow\left(t+1\right)\left(t^2+2t-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=\sqrt{6}-1>\sqrt{2}\left(l\right)\\t=\sqrt{6}+1>\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sinx+cosx=-1\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Rightarrow cos\left(x+\frac{\pi}{4}\right)=\pm\sqrt{1-sin^2\left(x+\frac{\pi}{4}\right)}=\pm\frac{\sqrt{2}}{2}\)