Biết rằng \(\dfrac{1}{2}\left[ {\cos \left( {\dfrac{\pi }{3} - 2x} \right) - \cos \left( {\dfrac{\pi }{2} + 2x} \right)} \right] \)\(- \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\) với mọi giá trị của góc lượng giác x ; trong đó a là số tự nhiên, b là số hữu tỉ thuộc \(\left[ {0;\dfrac{1}{2}} \right]\). Mệnh đề nào sau đây đúng?

Ta có: \(\dfrac{1}{2}\left[ {\cos \left( {\dfrac{\pi }{3} - 2x} \right) - \cos \left( {\dfrac{\pi }{2} + 2x} \right)} \right]\)\( - \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\)

\(\begin{array}{l} \Leftrightarrow \dfrac{1}{2}.\left( { - 2} \right).\sin \left( {\dfrac{{\dfrac{\pi }{3} - 2x + \dfrac{\pi }{2} + 2x}}{2}} \right).\\\sin \left( {\dfrac{{\dfrac{\pi }{3} - 2x - \dfrac{\pi }{2} - 2x}}{2}} \right) \\- \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\\ \Leftrightarrow  - \sin \dfrac{{5\pi }}{{12}}.\sin \left( { - \dfrac{\pi }{{12}} - 2x} \right) \\- \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\\ \Leftrightarrow \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{{12}}} \right).\sin \left( {\dfrac{\pi }{{12}} + 2x} \right)\\ - \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\\ \Leftrightarrow \cos \dfrac{\pi }{{12}}.\sin \left( {\dfrac{\pi }{{12}} + 2x} \right)\\ - \sin \dfrac{\pi }{{12}}.\cos \left( {\dfrac{\pi }{{12}} + 2x} \right) = \sin \left( {ax + b\pi } \right)\\ \Leftrightarrow \sin \left( {\dfrac{\pi }{{12}} + 2x - \dfrac{\pi }{{12}}} \right) = \sin \left( {ax + b\pi } \right) \\\Leftrightarrow \sin 2x = \sin \left( {ax + b\pi } \right)\\ \Rightarrow \left\{ \begin{array}{l}a = 2\\b = 2k\,\,\,\,\left( {k \in Z} \right)\end{array} \right.\,\,\,\,\,\,\\Do\,\,\,\,b \in \left[ {0;\dfrac{1}{2}} \right] \Rightarrow b = 0 \Rightarrow a + b = 2.\end{array}\)