Giải phương trình \({\sin ^3}x + {\cos ^3}x = 2\le...

C

- Hướng dẫn giải

Lời giải chi tiết:

\(\begin{array}{l}\,\,\,\,\,\,\,{\sin ^3}x + {\cos ^3}x = 2\left( {\sin x + \cos x} \right) - 1\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {1 - \sin x.\cos x} \right) - 2\left( {\sin x + \cos x} \right) + 1 = 0\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( { - 1 - \sin x.\cos x} \right) + 1 = 0\end{array}\)

Đặt \(\sin x + \cos x = t\,\,\,\,\left( { - \sqrt 2  \le t \le \sqrt 2 } \right)\), khi đó ta có:

\(\begin{array}{l}{\left( {\sin x + cosx} \right)^2} = {t^2} \Leftrightarrow {\sin ^2}x + 2\sin x.\cos x + {\cos ^2}x = {t^2}\\ \Leftrightarrow 1 + 2\sin x.\cos x = {t^2} \Leftrightarrow \sin x.\cos x = \dfrac{{{t^2} - 1}}{2}\end{array}\)
Phương trình trở thành:

\(\begin{array}{l} \Rightarrow t\left( { - 1 - \dfrac{{{t^2} - 1}}{2}} \right) + 1 = 0 \Leftrightarrow t\left( { - {t^2} - 1} \right) + 2 = 0 \Leftrightarrow  - {t^3} - t + 2 = 0 \Leftrightarrow t = 1\\ \Leftrightarrow \sin x + cosx = 1 \Leftrightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 1 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4} \Leftrightarrow \left[ \begin{array}{l}x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \dfrac{\pi }{2} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy \(S = \left\{ {k2\pi ,\,\,\dfrac{\pi }{2} + k2\pi \,,\,\,k \in \mathbb{Z}} \right\}\).