Câu hỏi
Giải phương trình: \(2\sin x - \sqrt 3 = 0\)
- A \(x = \frac{\pi }{3} + k\pi ,x = \frac{{2\pi }}{3} + k\pi ,k \in \mathbb{Z}.\)
- B \(x = \frac{\pi }{3} + k2\pi ,x = \frac{{2\pi }}{3} + k2\pi ,k \in \mathbb{Z}.\)
- C \(x = \frac{\pi }{6} + k\pi ,x = \frac{{2\pi }}{6} + k\pi ,k \in \mathbb{Z}.\)
- D \(x = \frac{\pi }{6} + k2\pi ,x = \frac{{2\pi }}{6} + k2\pi ,k \in \mathbb{Z}.\)
Lời giải chi tiết:
\(\begin{array}{l}2\sin x - \sqrt 3 = 0 \Leftrightarrow \sin \,x = \frac{{\sqrt 3 }}{2} \Leftrightarrow \sin \,x = \sin \frac{\pi }{3}\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = \pi - \frac{\pi }{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array}\)
Chọn B.
Câu hỏi trước
