$\begin{array}{l} 1)\sin \left( {3x + {{15}^o}} \right) = \dfrac{{ - 1}}{2}\\  \Leftrightarrow \sin \left( {3x + {{15}^o}} \right) = \sin \left( { - {{30}^o}} \right)\\  \Leftrightarrow \left[ \begin{array}{l} 3x + {15^o} =  - {30^o} + k{360^o}\\ 3x + {15^o} = {210^o} + k{360^o} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x =  - {15^o} + k{120^o}\\ x = {65^o} + k{120^o} \end{array} \right.\left( {k \in \mathbb{Z}} \right)\\ 2)\sin \left( {2x + \dfrac{\pi }{3}} \right) = \sin \left( {3x - \dfrac{\pi }{6}} \right)\\  \Leftrightarrow \left[ \begin{array}{l} 2x + \dfrac{\pi }{3} = 3x - \dfrac{\pi }{6} + k2\pi \\ 2x + \dfrac{\pi }{3} = \dfrac{{7\pi }}{6} - 3x + k2\pi  \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{2} - k2\pi \\ x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{5} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$  

`~rai~`

\(1.sin(3x+15^o)=-\dfrac{1}{2}\\\Leftrightarrow sin(3x+15^o)=sin(-30^o)\\\Leftrightarrow \left[\begin{array}{I}3x+15^o=-30^o+k360^o\\3x+15^o=180^o-(-30^o)+k360^o\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}3x=-45^o+k360^o\\3x=195^o+k360^o\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-15^o+k120^o\\x=65^o+k120^o.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy phương trình có các họ nghiệm là:}\\x=-15^o+k120^o;x=65^o+k120^o.(k\in\mathbb{Z})\\2.sin\left(2x+\dfrac{\pi}{3}\right)=sin\left(3x-\dfrac{\pi}{6}\right)\\\Leftrightarrow \left[\begin{array}{I}2x+\dfrac{\pi}{3}=3x-\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{3}=\pi-3x+\dfrac{\pi}{6}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{6}-k2\pi\\5x=\dfrac{5\pi}{6}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{6}-k2\pi\\x=\dfrac{\pi}{6}+k\dfrac{2\pi}{5}.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy phương trình có các họ nghiệm là:}\\x=\dfrac{\pi}{6}-k2\pi;x=\dfrac{\pi}{6}+k\dfrac{2\pi}{5}.(k\in\mathbb{Z})\)

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