Quay lại đáp án

$\cos \left[ {\pi \left( {{a^2} + 2a - \frac{1}{2}} \right)} \right] - \sin \left( {\pi {a^2}} \right)$=0 $ \Leftrightarrow \cos \left[ {\frac{\pi }{2} - \pi \left( {{a^2} + 2a} \right)} \right] = \sin \left( {\pi {a^2}} \right)$ $ \Leftrightarrow \sin \left[ {\pi \left( {{a^2} + 2a} \right)} \right] = \sin \left( {\pi {a^2}} \right)$$ \Leftrightarrow \begin{cases} \pi \left( a^2 + 2a \right) = \pi a^2 + k2 \pi \\ \pi a^2 + 2a = \pi - \pi a^2 + k2\pi \end{cases} $$ \Leftrightarrow \begin{cases} a = k \in \mathbb{Z} \\ 2a^2 + 2a - \left( 2k + 1 \right) = 0 \end{cases} $$\left( {\text{*}} \right)$Do $\begin{cases}\left( {\text{*}} \right) \\a{\text{ > }}0 \\k \in \mathbb{Z} \\\end{cases} $ suy ra $\min a = \frac{{\sqrt 3 - 1}}{2}$

$\cos \left[ {\pi \left( {{a^2} + 2a - \frac{1}{2}} \right)} \right] - \sin \left( {\pi {a^2}} \right)$=0 $ \Leftrightarrow \cos \left[ {\frac{\pi }{2} - \pi \left( {{a^2} + 2a} \right)} \right] = \sin \left( {\pi {a^2}} \right)$ $ \Leftrightarrow \sin \left[ {\pi \left( {{a^2} + 2a} \right)} \right] = \sin \left( {\pi {a^2}} \right)$$ \Leftrightarrow \left[ {\begin{matrix} \pi \left( a^2 + 2a \right) = \pi a^2 + k2 \pi \\ \pi a^2 + 2a = \pi - \pi a^2 + k2\pi \end{matrix}} \right. $$ \Leftrightarrow \left[ {\begin{matrix} a = k \in \mathbb{Z} \\ 2a^2 + 2a - \left( 2k + 1 \right) = 0 \end{matrix}} \right. $$\left( {\text{*}} \right)$Do $\begin{cases}\left( {\text{*}} \right) \\a{\text{ > }}0 \\k \in \mathbb{Z} \\\end{cases} $ suy ra $\min a = \frac{{\sqrt 3 - 1}}{2}$