Sin$^{3}$x – Cos$^{3}$x = 1

Đáp án:

 $\left[\begin{array}{l}x =\dfrac{\pi}{2} + k2\pi\\x = \pi+ k2\pi\end{array}\right. \quad (k\in \Bbb Z)$

Giải thích các bước giải:

$\sin^3x – \cos^3x = 1$

$\Leftrightarrow (\sin x – \cos x)(1 + \sin x\cos x) – 1 = 0$

Đặt $t = \sin x – \cos x\qquad (|t|\leq \sqrt2)$

$\Rightarrow t^2 = 1 – 2\sin x\cos x$

$\Rightarrow \dfrac{1 – t^2}{2}=\sin x\cos x$

Phương trình trở thành:

$t.\left(1 +\dfrac{1- t^2}{2}\right) – 1 = 0$

$\Leftrightarrow t(3 – t^2) – 2 = 0$

$\Leftrightarrow t^3 – 3t + 2 = 0$

$\Leftrightarrow \left[\begin{array}{l}t = 1 \qquad (nhận)\\t = -2\quad (loại)\end{array}\right.$

Với $t = 1$ ta được:

$\sin x – \cos x = 1$

$\Leftrightarrow \sqrt2\sin\left(x – \dfrac{\pi}{4}\right) = 1$

$\Leftrightarrow \sin\left(x – \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt2}$

$\Leftrightarrow \left[\begin{array}{l}x -\dfrac{\pi}{4} = \dfrac{\pi}{4} + k2\pi\\x – \dfrac{\pi}{4} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{2} + k2\pi\\x = \pi+ k2\pi\end{array}\right. \quad (k\in \Bbb Z)$