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

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