Đáp án:

 Phương trình có các họ nghiệm là:

a) ${x = \dfrac{\pi }{6} + k\pi \left( {k \in Z} \right);x = \dfrac{1}{2} + k\dfrac{\pi }{2}\left( {k \in Z} \right)}$

b) ${x = k2\pi \left( {k \in Z} \right);x =  \pm \dfrac{{5\pi }}{3} + k4\pi \left( {k \in Z} \right)}$

c) ${x = \dfrac{\pi }{3} + k2\pi \left( {k \in Z} \right);x = \dfrac{{2\pi }}{9} + k2\pi \left( {k \in Z} \right)}$

d) ${x =  - \dfrac{{29\pi }}{{24}} + k\pi \left( {k \in Z} \right);x = \dfrac{{25\pi }}{{48}} + k\dfrac{\pi }{2}\left( {k \in Z} \right)}$

Giải thích các bước giải:

$\begin{array}{l}
a)\cos \left( {x + \dfrac{\pi }{3}} \right)\sin \left( {2x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos \left( {x + \dfrac{\pi }{3}} \right) = 0\\
\sin \left( {2x - 1} \right) = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = \dfrac{\pi }{2} + k\pi \\
2x - 1 = k\pi 
\end{array} \right.\left( {k \in Z} \right) \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \left( {k \in Z} \right)\\
x = \dfrac{1}{2} + k\dfrac{\pi }{2}\left( {k \in Z} \right)
\end{array} \right.\\
b)\sin x + \sqrt 3 \sin \dfrac{x}{2} = 0\\
 \Leftrightarrow 2\sin \dfrac{x}{2}.\cos \dfrac{x}{2} + \sqrt 3 \sin \dfrac{x}{2} = 0\\
 \Leftrightarrow \sin \dfrac{x}{2}\left( {2\cos \dfrac{x}{2} + \sqrt 3 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \dfrac{x}{2} = 0\\
\cos \dfrac{x}{2} = \dfrac{{ - \sqrt 3 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{2} = k\pi \\
\dfrac{x}{2} =  \pm \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \left( {k \in Z} \right)\\
x =  \pm \dfrac{{5\pi }}{3} + k4\pi \left( {k \in Z} \right)
\end{array} \right.\\
c)\sin 2x = \cos \left( {\dfrac{\pi }{6} - x} \right)\\
 \Leftrightarrow \cos \left( {\dfrac{\pi }{2} - 2x} \right) = \cos \left( {\dfrac{\pi }{6} - x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{2} - 2x = \dfrac{\pi }{6} - x + k2\pi \\
\dfrac{\pi }{2} - 2x =  - \dfrac{\pi }{6} + x + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \left( {k \in Z} \right)\\
x = \dfrac{{2\pi }}{9} + k2\pi \left( {k \in Z} \right)
\end{array} \right.\\
d)\cos \left( {x - \dfrac{{9\pi }}{4}} \right) = \sin \left( {3x + \dfrac{{2\pi }}{3}} \right)\\
 \Leftrightarrow \cos \left( {x - \dfrac{{9\pi }}{4}} \right) = \cos \left( {3x + \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{{9\pi }}{4} = 3x + \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{{9\pi }}{4} =  - 3x - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{{29\pi }}{{24}} + k\pi \left( {k \in Z} \right)\\
x = \dfrac{{25\pi }}{{48}} + k\dfrac{\pi }{2}\left( {k \in Z} \right)
\end{array} \right.
\end{array}$