2 câu trả lời
$$\eqalign{ & \cos a - {\mathop{\rm sina}\nolimits} = {1 \over 2} \Leftrightarrow {\mathop{\rm cosa}\nolimits} = \sin a + {1 \over 2} \cr & Ta\,\,co:\,\,{\sin ^2}a + {\cos ^2}a = 1\,\,\forall a \cr & \Rightarrow {\sin ^2}a + {\left( {\sin a + {1 \over 2}} \right)^2} = 1 \cr & \Leftrightarrow {\sin ^2}a + {\sin ^2}a + \sin a + {1 \over 4} = 1 \cr & \Leftrightarrow 2{\sin ^2}a + {\mathop{\rm sina}\nolimits} - {3 \over 4} = 0 \cr & \Leftrightarrow \left[ \matrix{ \sin a = {{ - 1 + \sqrt 7 } \over 2}\,\,\left( {tm} \right) \Rightarrow \cos a = {{1 + \sqrt 7 } \over 4} \hfill \cr \sin a = {{ - 1 - \sqrt 7 } \over 4}\,\,\left( {tm} \right) \Rightarrow \cos a = {{1 - \sqrt 7 } \over 4} \hfill \cr} \right. \cr} $$
\[\begin{array}{l} \cos a - \sin a = \frac{1}{2}\\ \Leftrightarrow \sqrt 2 \left( {\frac{{\sqrt 2 }}{2}\cos a - \frac{{\sqrt 2 }}{2}\sin a} \right) = \frac{1}{2}\\ \Leftrightarrow \sqrt 2 \cos \left( {a + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {a + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{4} = \cos \beta \\ \Leftrightarrow \left[ \begin{array}{l} a + \frac{\pi }{4} = \beta + k2\pi \\ a + \frac{\pi }{4} = - \beta + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} a = \beta - \frac{\pi }{4} + k2\pi \\ a = - \beta - \frac{\pi }{4} + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right). \end{array}\]
