Giải phương trình  \(\dfrac{1}{{\cos x}} - \dfrac{...

A

- Hướng dẫn giải

Lời giải chi tiết:

\(\dfrac{1}{{\cos x}} - \dfrac{1}{{\sin x}} = 2\sqrt 2 .\cos \left( {x + \dfrac{\pi }{4}} \right)\,\,\,\,\left( 1 \right)\)

ĐK: \(\left\{ \begin{array}{l}\cos x \ne 0\\\sin x \ne 0\end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow x \ne \frac{{k\pi }}{2}\,\,\left( {k \in \mathbb{Z}} \right)\).

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

Vậy \(S = \left\{ { \dfrac{{ \pi }}{4} + \dfrac{k\pi}{2}  ,\,\,k \in \mathbb{Z}} \right\}\).

Chọn A.