588. $y=x+\cos 2x$ in the interval $(0, \pi / 4)$.
Solution. $1^{\circ}$. We have $y^{\prime}=1-2 \sin 2 x$.

$2^{0}$. Determine the critical points belonging to the given interval: $1-2 \sin 2 x=0, \sin 2 x=1 / 2, 2 x=\pi / 6, x=\pi / 12$.

$3^{\circ}$. Find $y^{\prime \prime}=-4 \cos 2 x$.

$4^{0}$. Since $y^{\prime \prime}(\pi / 12)=-4 \cos (\pi / 6)=-2 \sqrt{3}<0$, the function has a maximum at $x=\pi / 12$:

$$
y_{\max }=y\left(\frac{\pi}{12}\right)=\frac{\pi}{12}+\frac{\sqrt{3}}{2} \approx 1.13
$$