Sin$^{3}$x – Cos$^{3}$x = 1

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Sin$^{3}$x – Cos$^{3}$x = 1

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  2. 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:

    $\sin x – \cos x = 1$

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

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

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

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

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