3. Ta có:
$\cos\left[\pi(x^2+2x-\dfrac{1}{2})\right]=\sin(\pi x^2)$
$\Leftrightarrow \sin[\pi(x^2+2x)]=\sin(\pi x^2)$
$\Leftrightarrow \left[\begin{array}{l}\pi(x^2+2x)=\pi x^2+k2\pi\\\pi(x^2+2x)=\pi-\pi x^2+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x^2+2x=x^2+2k\\x^2+2x=1-x^2+2k\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=k\\2x^2+2x-(2k+1)=0\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=k\\x=\dfrac{1}{2}(-\sqrt{4k+3}-1)\\x=\dfrac{1}{2}(\sqrt{4k+3}-1)\end{array}\right.,k\in\mathbb{Z}$
Vậy nghiệm dương nhỏ nhất của phương trình đã cho là: $x=\dfrac{1}{2}(\sqrt3-1)$