Đáp án:
$\begin{array}{l}
a)\sin \left( {x + 2} \right) = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow \sin \left( {x + 2} \right) = \sin \left( {\dfrac{\pi }{3}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = \dfrac{\pi }{3} + k2\pi \\
x + 2 = \pi - \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} - 2 + k2\pi \\
x = \dfrac{{2\pi }}{3} - 2 + k2\pi
\end{array} \right.\\
Vậy\,x = \dfrac{\pi }{3} - 2 + k2\pi ;x = \dfrac{{2\pi }}{3} - 2 + k2\pi \\
b)\sin 3x - \cos 5x = 0\\
\Leftrightarrow \sin 3x = \cos 5x\\
\Leftrightarrow \sin 3x = \sin \left( {\dfrac{\pi }{2} - 5x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} - 5x + k2\pi \\
3x = \pi - \dfrac{\pi }{2} + 5x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{4}\\
x = \dfrac{{ - \pi }}{4} - k\pi
\end{array} \right.\\
Vậy\,x = \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{4};x = \dfrac{{ - \pi }}{4} - k\pi \\
c)\cos \left( {\dfrac{{2x}}{3} - \dfrac{x}{3}} \right) = 1\\
\Leftrightarrow \cos \dfrac{x}{3} = 1\\
\Leftrightarrow \dfrac{x}{3} = k2\pi \\
\Leftrightarrow x = k6\pi \\
Vậy\,x = k6\pi
\end{array}$
