Chứng minh: ${\sin ^2}a - {\sin ^2}b = \sin \left( {a + b} \right)\sin \left( {a - b} \right)$
$\begin{array}{l} \sin \left( {a + b} \right)\sin \left( {a - b} \right)\\ = \left( {\sin a\cos b + \sin b\cos a} \right)\left( {\sin a\cos b - \sin b\cos a} \right)\\ = {\left( {\sin a\cos b} \right)^2} - {\left( {\cos a\sin b} \right)^2}\\ = {\sin ^2}a\left( {1 - {{\sin }^2}b} \right) - \left( {1 - {{\sin }^2}a} \right){\sin ^2}b\\ = {\sin ^2}a - {\sin ^2}b\left( {dpcm} \right) \end{array}$
$\begin{array}{l}
{\sin ^2}\left( {x + {{45}^o}} \right) - {\sin ^2}\left( {x - {{30}^o}} \right) = \sin {15^o}\cos \left( {2x + {{15}^o}} \right)\\
\Leftrightarrow \sin \left( {x + {{45}^o} + x - {{30}^o}} \right)\sin \left( {x + {{45}^o} - x + {{30}^o}} \right) = \sin {15^o}\cos \left( {2x + {{15}^o}} \right)\\
\Leftrightarrow \sin \left( {2x + {{15}^o}} \right)\sin {75^o} = \sin {15^o}\cos {\left( {2x + 15} \right)^o}\\
\Leftrightarrow \sin \left( {2x + {{15}^o}} \right)\cos {15^o} - \sin {15^o}\cos {\left( {2x + 15} \right)^o} = 0\\
\Leftrightarrow \sin \left( {2x + {{15}^o} - {{15}^o}} \right) = 0\\
\Leftrightarrow \sin 2x = 0\\
\Leftrightarrow 2x = k{180^o}\\
\Leftrightarrow x = k{90^o}\\
x \in \left( { - {{90}^o};360} \right) \Rightarrow k = 0,1,2,3\\
\Rightarrow x = {0^o},{90^o},{180^o},{270^o}
\end{array}$
