$y= \sin^2\left(\dfrac{\pi}{2}-2x\right)+\dfrac{\pi}{2}x-\dfrac{\pi}{8}\\ y'=2\sin\left(\dfrac{\pi}{2}-2x\right)\cos\left(\dfrac{\pi}{2}-2x\right).(-2)+\dfrac{\pi}{2}\\ =-2\sin\left(\pi-4x\right)+\dfrac{\pi}{2}\\ y'=\dfrac{\pi}{2}\\ \Leftrightarrow -2\sin\left(\pi-4x\right)+\dfrac{\pi}{2}=\dfrac{\pi}{2}\\ \Leftrightarrow -2\sin\left(\pi-4x\right)=0\\ \Leftrightarrow \sin\left(\pi-4x\right)=0\\ \Leftrightarrow \pi-4x=k\pi(k \in \mathbb{Z})\\ \Leftrightarrow x=\dfrac{\pi -k\pi}{4}(k \in \mathbb{Z})\\ x \in (0;\pi) \Leftrightarrow 0<\dfrac{\pi -k\pi}{4}<\pi\\ \Leftrightarrow 0<\pi(1 -k)<4pi\\ \Leftrightarrow 0<1 -k<4\\ \Leftrightarrow \left\{\begin{array}{l} 1-k>0\\ 1-k<4\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} k<1\\ k>-3\end{array} \right.\\ \Leftrightarrow k \in \{-2;-1;0\}\\ \Rightarrow x\in\left\{\dfrac{3\pi}{4};\dfrac{\pi}{2};\dfrac{\pi}{4}\right\}$
