Câu hỏi:
Giải phương trình \({\mathop{\rm s}\nolimits} {\rm{in3}}x + \sqrt 3 \cos 3x = 2\sin x\)
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A.
\(\left[ \begin{array}{l}x = – \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{{5\pi }}{{12}} + k\pi \end{array} \right.\) -
B.
\(\left[ \begin{array}{l}x = – \dfrac{\pi }{6} + k\pi \\x = \dfrac{\pi }{6} + k\dfrac{\pi }{2}\end{array} \right.\) -
C.
\(\left[ \begin{array}{l}x = – \dfrac{\pi }{6} + k\pi \\x = \dfrac{\pi }{3} + k\pi \end{array} \right.\) -
D.
\(\left[ \begin{array}{l}x = – \dfrac{\pi }{{12}} + k\pi \\x = \dfrac{{5\pi }}{{24}} + k\dfrac{\pi }{2}\end{array} \right.\)
Lời giải tham khảo:
Đáp án đúng: B
Ta có: \({\mathop{\rm s}\nolimits} {\rm{in3}}x + \sqrt 3 \cos 3x = 2\sin x \) \( \Leftrightarrow 2\left( {\frac{1}{2}\sin 3x + \frac{{\sqrt 3 }}{2}\cos 3x} \right) = 2\sin x\) \(\Leftrightarrow 2\sin \left( {3x + \dfrac{\pi }{3}} \right) = 2\sin x\)
\( \Leftrightarrow \sin \left( {3x + \dfrac{\pi }{3}} \right) = \sin x\) \( \Leftrightarrow \left[ \begin{array}{l}3x + \dfrac{\pi }{3} = x + k2\pi \\3x + \dfrac{\pi }{3} = \pi – x + k2\pi \end{array} \right. \) \(\Leftrightarrow \left[ \begin{array}{l}x = – \dfrac{\pi }{6} + k\pi \\x = \dfrac{\pi }{6} + k\dfrac{\pi }{2}\end{array} \right.\;\left( {k \in \mathbb{Z}} \right)\)
Chọn đáp án B.
Trả lời