Sửa đề: \(sin^3x+cos^3x-2\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)+1=0\)
\(\Leftrightarrow sin^3x+cos^3x-2\left(cosx+sinx\right)+1=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x-sinx.cosx+cos^2x\right)-2\left(sinx+cosx\right)+1=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)-2\left(sinx+cosx\right)+1=0\)
\(\Leftrightarrow sinx-sin^2x.cosx+cosx-sinx.cos^2x-2\left(sinx+cosx\right)+1=0\)
\(\Leftrightarrow-sinx-sin^2x.cosx-cosx-sinx.cos^2x+1=0\)
\(\Leftrightarrow-\left(sinx+cosx\right)\left(1+sinx.cosx\right)+1=0\)
Đặt \(t=sinx+cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(\Rightarrow\dfrac{t^2-1}{2}=sinx.cosx\)
Pttt: \(-t\left(1+\dfrac{t^2-1}{2}\right)+1=0\)
\(\Leftrightarrow-t+t^3+2=0\)
\(\Leftrightarrow t=1\)
\(\Rightarrow sinx+cosx=1\)\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
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