$y = \sqrt{3}\sin 4x + 2\sin^2 2x - 1\\ y = \sqrt{3}\sin 4x - \cos 4x\\ y = 2\left(\dfrac{\sqrt{3}}{2}\sin 4x - \dfrac{1}{2}\cos 4x\right)\\ y = 2\sin\left(4x - \dfrac{\pi}{6}\right)$$\\$ Do $-1 \leq \sin\left(4x - \dfrac{\pi}{6}\right) \leq 1$$\\$ $\Rightarrow -2 \leq 2\sin\left(4x - \dfrac{\pi}{6}\right) \leq 2\\ \Rightarrow y \geq -2$$\\$ Dấu "=" xảy ra khi $\sin\left(4x - \dfrac{\pi}{6}\right) = -1$ $\Rightarrow 4x - \dfrac{\pi}{6} = -\dfrac{\pi}{2} + k2\pi\ (k \in \mathbb{Z})\\\Rightarrow 4x = -\dfrac{\pi}{3} + k2\pi \ (k \in \mathbb{Z})\\ \Rightarrow x = -\dfrac{\pi}{12} + \dfrac{k\pi}{2}\ (k \in \mathbb{Z})$$\\$ Vậy giá trị nhỏ nhất của hàm số là $-2$ khi $x = -\dfrac{\pi}{12} + \dfrac{k\pi}{2}\ (k \in \mathbb{Z})$