\(cos^3x-sin^3x=cos2x\)
\(\Leftrightarrow\left(cosx-sinx\right).\left(1+cosx.sinx\right)=cos^2x-sin^2x\)
\(\Leftrightarrow\left(cosx-sinx\right).\left[\left(1+cosx.sinx\right)-\left(cosx+sinx\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\left(1\right)\\1+cosx.sinx-\left(cosx+sinx\right)=0\left(2\right)\end{matrix}\right.\)
(1): \(cosx-sinx=0\)
\(\Leftrightarrow tanx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)
(2): \(1+cosx.sinx-\left(cosx+sinx\right)=0\)
Đặt \(cosx+sinx=t,t\in\left[-\sqrt{2},\sqrt{2}\right]\)
\(\rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)
\(pt\Leftrightarrow1+\dfrac{1^2-1}{2}-t=0\)
\(\Leftrightarrow2+t^2-1-2t=0\)
\(\Leftrightarrow t^2-2t+1=0\)
\(\Leftrightarrow t=1\left(tm\right)\)
Với t = 1 \(\Rightarrow cosx+sinx=1\)
\(\Leftrightarrow\sqrt{2}.sin.\left(x-\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin.\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow sin.\left(x-\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
