Đáp án:
$\left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x = k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin^3x +\cos^3x = \sin x +\cos x$
$\to (\sin x + \cos x)(\sin^2x - \sin \cos x +\cos^2x) = \sin x +\cos x$
$\to (\sin x+\cos x)(1-\sin x\cos x - 1) = 0$
$\to (\sin x +\cos x)\sin x\cos x = 0$
$\to \sin\left(x +\dfrac{\pi}{4}\right).\sin2x = 0$
$\to \left[\begin{array}{l}\sin\left(x +\dfrac{\pi}{4}\right) = 0\\\sin2x = 0\end{array}\right.$
$\to \left[\begin{array}{l}x +\dfrac{\pi}{4} = k\pi\\2x = k\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x = k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)$
