3. The minimum value of the function $y=|\cos x|+|\cos 2 x|(x \in \mathbf{R})$ is $\qquad$
3. $\frac{\sqrt{2}}{2}$ Detailed Explanation: Let $t=|\cos x|$, then $0 \leqslant t \leqslant 1, y=t+\left|2 t^{2}-1\right|$. When $0 \leqslant t<\frac{\sqrt{2}}{2}$, $y=-2\left(t-\frac{1}{4}\right)^{2}+\frac{9}{8}$, we get $\frac{\sqrt{2}}{2}<y \leqslant \frac{9}{8}$; When $\frac{\sqrt{2}}{2} \leqslant t \leqslant 1$, $y=2\left(t+\frac{1}{4}\right)^{2}-\frac{9}{8}$, we get $\frac{\sqrt{2}}{2} \leqslant y \leqslant 2$. Therefore, $y_{\text {min }}=\frac{\sqrt{2}}{2}$.