Phương trình: \(2\sqrt 3 \sin \left( {x - \frac{\p...

A

- Hướng dẫn giải

Phương pháp giải:

Dùng hai công thức biến đổi : \(\sin 2x = 2\sin x\cos x;\,\,2{\cos ^2}x = 1 + \cos 2x.\)

Giải chi tiết:

\(\begin{array}{l}\,\,\,\,\,\,\,2\sqrt 3 \sin \left( {x - \frac{\pi }{8}} \right)\cos \left( {x - \frac{\pi }{8}} \right) + 2{\cos ^2}\left( {x - \frac{\pi }{8}} \right) = \sqrt 3  + 1\\ \Leftrightarrow \sqrt 3 \sin \left( {2x - \frac{\pi }{4}} \right) + 1 + \cos \left( {2x - \frac{\pi }{4}} \right) = \sqrt 3  + 1\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{4}} \right).\frac{{\sqrt 3 }}{2} + \cos \left( {2x - \frac{\pi }{4}} \right).\frac{1}{2} = \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{4}} \right).\cos \frac{\pi }{6} + \cos \left( {2x - \frac{\pi }{4}} \right).\sin \frac{\pi }{6} = \sin \frac{\pi }{3}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{{12}}} \right) = \sin \frac{\pi }{3}\\ \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{{12}} = \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{{12}} = \pi  - \frac{\pi }{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = \frac{{5\pi }}{{12}} + k2\pi \\2x = \frac{{3\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{24}} + k\pi \\x = \frac{{3\pi }}{8} + k\pi \end{array} \right.,k \in \mathbb{Z}.\end{array}\)

Chọn A.