Giải phương trình \(\sin \left( {\dfrac{\pi }{3} - 2x} \right) = \sin \left( {x + \dfrac{\pi }{2}} \right)\) ta được:
A.\(\left[ \begin{array}{l}x = - \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\x = - \dfrac{\pi }{6} + k2\pi \end{array} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)
B.\(\left[ \begin{array}{l}x = - \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3}\\x = - \dfrac{\pi }{6} + k\pi \end{array} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)
C.\(x = - \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\) \(\left( {k \in \mathbb{Z}} \right)\)
D.\(x = - \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3}\)\(\left( {k \in \mathbb{Z}} \right)\)
A.\(\left[ \begin{array}{l}x = - \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\\x = - \dfrac{\pi }{6} + k2\pi \end{array} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)
B.\(\left[ \begin{array}{l}x = - \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3}\\x = - \dfrac{\pi }{6} + k\pi \end{array} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)
C.\(x = - \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{3}\) \(\left( {k \in \mathbb{Z}} \right)\)
D.\(x = - \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3}\)\(\left( {k \in \mathbb{Z}} \right)\)
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Hóa Học lớp 12
