A. \(\left[ \begin{array}{l}x = k\pi \\x = \dfrac{\pi }{{20}} + k\dfrac{\pi }{{10}}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).
B. \(\left[ \begin{array}{l}x = k\dfrac{\pi }{2}\\x = \dfrac{\pi }{{20}} + k\dfrac{\pi }{{10}}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).
C. \(\left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\dfrac{\pi }{{10}}\\x = \dfrac{\pi }{{20}} + k\dfrac{\pi }{{10}}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).
D. \(\left[ \begin{array}{l}x = k\dfrac{\pi }{2}\\x = \dfrac{\pi }{{20}} + k\dfrac{\pi }{5}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).
B
Ta có: \(\sin 5x.\cos 3x = \sin 7x.\cos 5x\)
\( \Leftrightarrow \dfrac{1}{2}\left( {\sin 8x + \sin 2x} \right) = \dfrac{1}{2}\left( {\sin 12x + \sin 2x} \right)\)
\( \Leftrightarrow \sin 8x = \sin 12x\)
\( \Leftrightarrow \left[ \begin{array}{l}12x = 8x + k2\pi \\12x = \pi - 8x + k2\pi \end{array} \right. \)
\( \Leftrightarrow \left[ \begin{array}{l}
4x = k2\pi \\
20x = \pi + k2\pi
\end{array} \right.\)
\(\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x = \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{{10}}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
Chọn đáp án B.
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