Đáp án:
$\begin{array}{l}
1)\cos \left( {\dfrac{{3x}}{5} - \dfrac{\pi }{3}} \right) = - \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{{3x}}{5} - \dfrac{\pi }{3} = \dfrac{{3\pi }}{4} + k2\pi \\
\dfrac{{3x}}{5} - \dfrac{\pi }{3} = - \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{65\pi }}{{36}} + \dfrac{{k10\pi }}{3}\\
x = - \dfrac{{25\pi }}{{36}} + \dfrac{{k10\pi }}{3}
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x = \dfrac{{65\pi }}{{36}} + \dfrac{{k10\pi }}{3}\\
x = - \dfrac{{25\pi }}{{36}} + \dfrac{{k10\pi }}{3}
\end{array} \right.\\
2)\cos \left( {x + 1} \right) = \dfrac{2}{3}\\
\Leftrightarrow x + 1 = \pm \arccos \dfrac{2}{3} + k2\pi \\
\Leftrightarrow x = - 1 \pm \arccos \dfrac{2}{3} + k2\pi \\
Vậy\,x = - 1 \pm \arccos \dfrac{2}{3} + k2\pi \\
3)\sin \left( {\dfrac{{2x}}{3} - \dfrac{\pi }{6}} \right) = 1\\
\Leftrightarrow \dfrac{{2x}}{3} - \dfrac{\pi }{6} = \dfrac{\pi }{2} + 2k\pi \\
\Leftrightarrow x = \pi + 3k\pi \\
Vậy\,x = \pi + 3k\pi
\end{array}$
