\(sin^3x+cos^3x=sinx-cosx\)
TH1 cosx=0 và sinx=1 \(\Leftrightarrow1=1\) (luôn đúng)
\(\Rightarrow x=\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)
TH2 chia 2 về cho \(cos^3x\)
\(\Leftrightarrow tan^3x+1=tanx\left(1+tan^2x\right)-\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^2x-tanx+2=0\left(vl\right)\)
vậy \(x=\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)