\(\Leftrightarrow\left(cos^3x-sin^3x\right)+\left(1-sin2x\right)=0\)
\(\Leftrightarrow\left(cos^3x-sin^3x\right)+\left(sin^2x+cos^2x-2sinx.cosx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(1+sinx.cosx\right)+\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(1+sinx.cosx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\Leftrightarrow x=\frac{\pi}{4}+k\pi\\1+sinx.cosx+cosx-sinx=0\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)
Pt trở thành: \(1+\frac{1-t^2}{2}-t=0\)
\(\Leftrightarrow-t^2-2t+3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
