Ta có $-1\le \cos x\le 1$ nên
$\Rightarrow 2\le 3+\cos x\le 4$
$\Rightarrow \dfrac 1 2\le y\le 1$
$\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} \min y = \dfrac{1}{2} \Rightarrow \cos x = 1\\ \max y = 1 \Rightarrow \cos x = - 1 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \min y = \frac{1}{2} \Rightarrow x = k2\pi \\ \max y = 1 \Rightarrow x = \pi + k2\pi \end{array} \right. \end{array}$