Đáp án:

$\begin{array}{l}
a)\sin \left( {{{15}^0} - 2x} \right) =  - 1\\
 \Rightarrow {15^0} - 2x =  - {90^0} + k{.360^0}\\
 \Rightarrow 2x = {105^0} - k{.360^0}\\
 \Rightarrow x = 52,{5^0} + k{.180^0}\\
b)\sin 2x + \sin 5x = 0\\
 \Rightarrow sin2x =  - sin5x\\
 \Rightarrow sin2x = \sin \left( { - 5x} \right)\\
 \Rightarrow \left[ \begin{array}{l}
2x =  - 5x + k2\pi \\
2x = \pi  + 5x + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{7}\\
x = \dfrac{{ - \pi }}{3} - \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)\\
c)\cos \left( {x - {3^0}} \right) = \dfrac{{\sqrt 2 }}{2}\\
 \Rightarrow \cos \left( {x - {3^0}} \right) = \cos {45^0}\\
 \Rightarrow \left[ \begin{array}{l}
x - {3^0} = {45^0} + k{.360^0}\\
x - {3^0} =  - {45^0} + k{.360^0}
\end{array} \right.\left( {k \in Z} \right)\\
 \Rightarrow \left[ \begin{array}{l}
x = {48^0} + k{.360^0}\\
x =  - {42^0} + k{.360^0}
\end{array} \right.\left( {k \in Z} \right)\\
d){\sin ^2}x + {\sin ^2}3x = 1\\
 \Rightarrow {\sin ^2}x + {\sin ^2}3x = {\sin ^2}x + {\cos ^2}x\\
 \Rightarrow {\sin ^2}3x = {\cos ^2}x\\
 \Rightarrow \left[ \begin{array}{l}
\sin 3x = \cos x\\
\sin 3x =  - \cos x
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\sin 3x = \sin \left( {\dfrac{\pi }{2} - x} \right)\\
\sin 3x = \sin \left( {x - \dfrac{\pi }{2}} \right)
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{2} - x + k2\pi \\
3x = \pi  - \dfrac{\pi }{2} + x + k2\pi \\
3x = x - \dfrac{\pi }{2} + k2\pi \\
3x = \pi  - x + \dfrac{\pi }{2} + k2\pi 
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{4} + k\pi \\
x = \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$