Đáp án:
$\begin{array}{l}
a)\sin \left( {3x + \dfrac{\pi }{2}} \right) = \dfrac{1}{2}\\
\Leftrightarrow \sin \left( {3x + \dfrac{\pi }{2}} \right) = \sin \dfrac{\pi }{6}\\
\Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{2} = \dfrac{\pi }{6} + k2\pi \\
3x + \dfrac{\pi }{2} = \pi - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - \pi }}{9} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - \pi }}{9} + \dfrac{{k2\pi }}{3};x = \dfrac{\pi }{9} + \dfrac{{k2\pi }}{3}\\
b)\sin \left( {x - \dfrac{\pi }{5}} \right) = - 1\\
\Leftrightarrow x - \dfrac{\pi }{5} = \dfrac{{ - \pi }}{2} + k2\pi \\
\Leftrightarrow x = \dfrac{{ - 3\pi }}{{10}} + k2\pi \\
Vậy\,x = \dfrac{{ - 3\pi }}{{10}} + k2\pi \\
c)\sin \left( {2x - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
2x - \dfrac{\pi }{4} = \pi - \dfrac{\pi }{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
Vậy\,x = \dfrac{\pi }{4} + k\pi ;x = \dfrac{\pi }{2} + k\pi
\end{array}$
