Nghiệm của phương trình \(\sin \left( {4x + \frac{1}{2}} \right) = \frac{1}{3}\) là:
A.A.
\(\left[ \begin{array}{l}
x = - \frac{1}{8} + k\frac{\pi }{2}\\
x = \frac{\pi }{4} + k\frac{\pi }{2}
\end{array} \right.,k \in Z\)
x = - \frac{1}{8} + k\frac{\pi }{2}\\
x = \frac{\pi }{4} + k\frac{\pi }{2}
\end{array} \right.,k \in Z\)
B.B.
\(\left[ \begin{array}{l}
x = - \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}\\
x = \frac{\pi }{4} - \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}
\end{array} \right.,k \in Z\)
x = - \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}\\
x = \frac{\pi }{4} - \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}
\end{array} \right.,k \in Z\)
C.C.
\(\left[ \begin{array}{l}
x = \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}\\
x = \frac{\pi }{4} - \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}
\end{array} \right.,k \in Z\)
x = \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}\\
x = \frac{\pi }{4} - \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}
\end{array} \right.,k \in Z\)
D.D.
\(\left[ \begin{array}{l}
x = - \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}\\
x = \frac{\pi }{4} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}
\end{array} \right.,k \in Z\)
x = - \frac{1}{8} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}\\
x = \frac{\pi }{4} - \frac{1}{4}\arcsin \frac{1}{3} + k\frac{\pi }{2}
\end{array} \right.,k \in Z\)
Đáp án và lời giải
Đáp án:B