Câu hỏi :

Tìm nghiệm của phương trình \(\sin x + \sqrt 3 \cos x = \sqrt 2 \).

A. \(x = - \dfrac{\pi }{{12}} + k2\pi ,\;x = \dfrac{{5\pi }}{{12}} + k2\pi ,\;\left( {k \in \mathbb{Z}} \right).\)

B. \(x = - \dfrac{\pi }{4} + k2\pi ,\;x = \dfrac{{3\pi }}{4} + k2\pi ,\;\left( {k \in \mathbb{Z}} \right).\)

C. \(x = \dfrac{\pi }{3} + k2\pi ,\;x = \dfrac{{2\pi }}{3} + k2\pi ,\;\left( {k \in \mathbb{Z}} \right).\)

D. \(x = - \dfrac{\pi }{4} + k2\pi ,\;x = - \dfrac{{5\pi }}{4} + k2\pi ,\;\left( {k \in \mathbb{Z}} \right).\)

* Đáp án

A

* Hướng dẫn giải

Ta có: \(\sin x + \sqrt 3 \cos x = \sqrt 2 \)

\(\begin{array}{l} \Leftrightarrow \frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x = \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \cos \frac{\pi }{3}\sin x + \sin \frac{\pi }{3}\cos x = \frac{{\sqrt 2 }}{2} \end{array}\)

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

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