Giải phương trình \(\cos x\sin x + \left| {\cos x...

B

- Hướng dẫn giải

Lời giải chi tiết:

\(\cos x\sin x + \left| {\cos x + \sin x} \right| = 1 \Leftrightarrow \cos x\sin x + \left| {\sin x + \cos x} \right| = 1\,\,\left( 1 \right)\)

Đặt \(\left| {\sin x + \cos x} \right| = t\,\,\,\,\left( {0 \le t \le \sqrt 2 } \right)\), khi đó ta có:

\({\left( {\sin x + \cos x} \right)^2} = {t^2} \Leftrightarrow 1 + 2\sin x.cosx = {t^2} \Leftrightarrow \sin x.cosx = \dfrac{{{t^2} - 1}}{2}\)

Khi đó phương trình trở thành: \(\dfrac{{{t^2} - 1}}{2} + t = 1 \Leftrightarrow {t^2} + 2t - 3 = 0 \Leftrightarrow \left[ \begin{array}{l}t = 1\,\,\,\,\,\,\,\left( {tm} \right)\\t =  - 3\,\,\left( {ktm} \right)\end{array} \right.\)

\(\begin{array}{l} \Rightarrow \left| {\sin x + \cos x} \right| = 1 \Leftrightarrow \left[ \begin{array}{l}\sin x + \cos x = 1\\\sin x + \cos x =  - 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) =  - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\\\sin \left( {x + \dfrac{\pi }{4}} \right) =  - \dfrac{1}{{\sqrt 2 }}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi \\x + \dfrac{\pi }{4} =  - \dfrac{\pi }{4} + k2\pi \\x + \dfrac{\pi }{4} = \dfrac{{5\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \dfrac{\pi }{2} + k2\pi \\x =  - \dfrac{\pi }{2} + k2\pi \\x = \pi  + k2\pi \end{array} \right. \Leftrightarrow x = \dfrac{{k\pi }}{2}\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

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