2 câu trả lời
\[\begin{array}{l} \sin \left( {4x + 1} \right) = \frac{1}{3}\\ \Leftrightarrow \left[ \begin{array}{l} 4x + 1 = \arcsin \frac{1}{3} + k2\pi \\ 4x + 1 = \pi - \arcsin \frac{1}{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} 4x = \arcsin \frac{1}{3} - 1 + k2\pi \\ 4x = \pi - \arcsin \frac{1}{3} - 1 + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{1}{4}\left( {\arcsin \frac{1}{3} - 1 + k2\pi } \right)\\ x = \frac{1}{4}\left( {\pi - \arcsin \frac{1}{3} - 1 + k2\pi } \right) \end{array} \right.. \end{array}\]
Lời giải: \(\eqalign{ & \sin \left( {4x + 1} \right) = {1 \over 3} \cr & \Leftrightarrow \left[ \matrix{ 4x + 1 = \arcsin {1 \over 3} + k2\pi \hfill \cr 4x + 1 = \pi - \arcsin {1 \over 3} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ 4x = - 1 + \arcsin {1 \over 3} + k2\pi \hfill \cr 4x = - 1 + \pi - \arcsin {1 \over 3} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = - {1 \over 4} + {1 \over 4}\arcsin {1 \over 3} + {{k\pi } \over 2} \hfill \cr x = - {1 \over 4} + {\pi \over 4} - {1 \over 4}\arcsin {1 \over 3} + {{k\pi } \over 2} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} \)
