$\begin{array}{l}
\cos \left( {x + {{120}^o}} \right) - \sin \left( {2x + {{50}^o}} \right) = 0\\
\Leftrightarrow \cos \left( {x + {{120}^o}} \right) = \sin \left( {2x + {{50}^o}} \right)\\
\Leftrightarrow \cos \left[ {{{90}^o} - \left( { - x - {{30}^o}} \right)} \right] = \sin \left( {2x + {{50}^o}} \right)\\
\Leftrightarrow \sin \left( { - x - {{30}^o}} \right) = \sin \left( {2x + {{50}^o}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
- x - {30^o} = 2x + {50^o} + k{360^o}\\
- x - {30^o} = {180^o} - 2x - {50^o} + k{360^o}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = - {80^o} - k{360^o}\\
x = {160^o} + k{360^o}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - {{80}^o}}}{3} - k{120^o}\\
x = {160^o} + k{360^o}
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$
