Tìm gtln gtnn của y=x-sin2x x thuộc [-pi;pi]

$y'=1-2cos2x$

$y'=0 ↔ cos2x=\dfrac{1}{2} ↔ \left[ \begin{array}{l}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{array} \right.$

$↔ \left[ \begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{array} \right.$

Vì $x∈[-\pi;\pi]$ nên $x∈\Bigg\{-\dfrac{5\pi}{6};-\dfrac{\pi}{6};\dfrac{\pi}{6};\dfrac{5\pi}{6}\Bigg\}$

Dựa vào bảng biến thiên suy ra:

GTLN là: $y\Bigg(\dfrac{5\pi}{6}\Bigg)=\dfrac{5\pi+3\sqrt[]{3}}{6}$

 GTNN là: $y\Bigg(-\dfrac{5\pi}{6}\Bigg)=-\dfrac{5\pi+3\sqrt[]{3}}{6}$