Nghiệm của phương trình \(\sin 2x - \cos x = 0\) là:
-
A.
\(\left[ \begin{array}{l}x = - \frac{\pi }{6} + \frac{{k2\pi }}{3}\\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\) .
-
B.
\(\left[ \begin{array}{l}x = \frac{\pi }{6} + \frac{{k2\pi }}{3}\\x = \frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\).
-
C.
\(\left[ \begin{array}{l}x = - \frac{\pi }{6} + \frac{{k2\pi }}{3}\\x = \frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\).
-
D.
\(\left[ \begin{array}{l}x = - \frac{\pi }{6} + \frac{{k2\pi }}{3}\\x = \frac{\pi }{2} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\).
Phương trình \(\sin x = \sin \alpha \)có nghiệm: \(x = \alpha + k2\pi ,k \in \mathbb{Z}\) và \(x = \pi - \alpha + k2\pi ,k \in \mathbb{Z}\)
\(\sin 2x - \cos x = 0 \Leftrightarrow \sin 2x = \cos x \Leftrightarrow \sin 2x = \sin \left( {\frac{\pi }{2} - x} \right) \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{2} - x + k2\pi \\2x = \pi - \left( {\frac{\pi }{2} - x} \right) + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{2} + k2\pi \\x = \frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + \frac{{k2\pi }}{3}\\x = \frac{\pi }{2} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
Đáp án : B