Giải phương trình lượng giác \(\sin \left( {\dfrac{\pi }{3} - 3x} \right) = \sin \left( {x + \dfrac{\pi }{4}} \right)\) có nghiệm là:

\(m =  - 3\)

\(m =  - 2\)      

\(m = 0\)

\(m = 3\)

\(\left[ \begin{array}{l}x = \alpha  + k\pi \\x = \pi  - \alpha  + k\pi \end{array} \right.\left( {k \in Z} \right)\)    

\(\left[ \begin{array}{l}x = \alpha  + k2\pi \\x = \pi  - \alpha  + k2\pi \end{array} \right.\left( {k \in Z} \right)\)

\(\left[ \begin{array}{l}x = \alpha  + k2\pi \\x =  - \alpha  + k2\pi \end{array} \right.\left( {k \in Z} \right)\)      

\(\left[ \begin{array}{l}x = \alpha  + k\pi \\x =  - \alpha  + k\pi \end{array} \right.\left( {k \in Z} \right)\)

\(x =  - \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\)

\(x = \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\)

\(x =  - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\)

\(x =  - \pi  + k2\pi \left( {k \in Z} \right)\)

\(\sin x =  - 1 \Leftrightarrow x =  - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\)           

\(\sin x = 0 \Leftrightarrow x = k\pi \left( {k \in Z} \right)\)

\(\sin x = 0 \Leftrightarrow x = k2\pi \left( {k \in Z} \right)\)                     

\(\sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\) 

\(x = \dfrac{{5\pi }}{6} + k2\pi\)

\(x = \dfrac{\pi }{6}\)

\(x = \dfrac{{5\pi }}{6}\)

\(x = \dfrac{\pi }{3}\)

\(1\)     

\(0\)      

\(3\)

\(2\)

\(x = \dfrac{\pi }{2} + k2\pi \).

\(x = \dfrac{k\pi }{2}\).

\(x = k2\pi \).

\(x = \dfrac{\pi }{6} + k2\pi \).