Đáp án:
$\left[\begin{array}{l}x = k2\pi\\x =-\dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\cos^3x - \sin^3x + 1 = 0$
$(\cos x - \sin x)^3 - 3\cos x\sin x(\cos x - \sin x) + 1 = 0$
Đặt $t = \cos x - \sin x \quad (|t| \leq \sqrt2)$
$\Rightarrow t^2 = 1 - 2\sin xcos x$
$\Rightarrow \dfrac{1 - t^2}{2}=\sin x\cos x$
Phương trình trở thành:
$t^3 - 3t.\dfrac{1 - t^2}{2} + 1 = 0$
$\Leftrightarrow 5t^3 - 3t + 2 = 0$
$\Leftrightarrow (t+1)(5t^2 - 5t + 2) = 0$
$\Leftrightarrow t + 1 = 0$
$\Leftrightarrow t = -1$
Ta được:
$\cos x - \sin x = 1$
$\Leftrightarrow \sqrt2\cos\left(x +\dfrac{\pi}{4}\right) = 1$
$\Leftrightarrow \cos\left(x +\dfrac{\pi}{4}\right) = \dfrac{\sqrt2}{2}$
$\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = \dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} =-\dfrac{\pi}{4} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k2\pi\\x =-\dfrac{\pi}{2} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
