$\begin{array}{l}
y = 1 - \sin \left( {x + \pi } \right) + 2\cos \left( {\dfrac{{2\pi  + x}}{2}} \right)\\
 = 1 - \left( { - \sin x} \right) + 2\cos \left( {\pi  + \dfrac{x}{2}} \right)\\
 = 1 + \sin x - 2\cos \dfrac{x}{2}\\
y' = \left( {1 + \sin x - 2\cos \dfrac{x}{2}} \right)'\\
 = \left( 1 \right)' + \left( {\sin x} \right)' - \left( {2\cos \dfrac{x}{2}} \right)'\\
 = \cos x - \left[ {\left( 2 \right)'\cos \dfrac{x}{2} + \left( {\cos \dfrac{x}{2}} \right)'.2} \right]\\
 = \cos x - \left( {\cos \dfrac{x}{2}} \right)'.2\\
 = \cos x - \left( {\cos \dfrac{x}{2}} \right)'.2\\
 = \cos x - \left( { - \sin \dfrac{x}{2}} \right).2.\left( {\dfrac{x}{2}} \right)'\\
 = \cos x + \sin \dfrac{x}{2}\\
y' = 0 \Leftrightarrow \cos x + 2\sin \dfrac{x}{2} = 0\\
 \Leftrightarrow 1 - 2{\sin ^2}\dfrac{x}{2} + \sin \dfrac{x}{2} = 0\\
 \Leftrightarrow 2{\sin ^2}\dfrac{x}{2} - \sin \dfrac{x}{2} - 1 = 0\\
 \Leftrightarrow \left( {2\sin \dfrac{x}{2} + 1} \right)\left( {\sin \dfrac{x}{2} - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \dfrac{x}{2} =  - \dfrac{1}{2}\\
\sin \dfrac{x}{2} = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{2} =  - \dfrac{\pi }{6} + k2\pi \\
\dfrac{x}{2} = \dfrac{{7\pi }}{6} + k2\pi \\
\dfrac{x}{2} = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{3} + k4\pi \\
x = \dfrac{{7\pi }}{3} + k4\pi \\
x = k2\pi 
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$

Đáp án:

\(S = \left\{\pi + k4\pi;\ -\dfrac{\pi}{4} + k4\pi;\ \dfrac{7\pi}{3} + k4\pi\ \Bigg|\ k\in\Bbb Z\right\}\) 

Giải thích các bước giải:

\(\begin{array}{l}
\quad y = 1  -\sin(x+\pi) + 2\cos\left(\dfrac{2\pi + x}{2}\right)\\
\Leftrightarrow y = 1 + \sin x - 2\cos\dfrac x2\\
\Rightarrow y' = \cos x + \sin\dfrac x2\\
\text{Khi đó:}\\
\quad y' =0\\
\Leftrightarrow \cos x + \sin\dfrac x2 = 0\\
\Leftrightarrow 1 - 2\sin^2\dfrac x2 + \sin\dfrac x2 = 0\\
\Leftrightarrow \left(\sin\dfrac x2 -1\right)\left(2\sin\dfrac x2 + 1\right) = 0\\
\Leftrightarrow \left[\begin{array}{l}\sin\dfrac x2 = 1\\\sin\dfrac x2 = - \dfrac12\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\dfrac x2 = \dfrac{\pi}{2} + k2\pi\\\dfrac x2 = - \dfrac{\pi}{6} + k2\pi\\\dfrac x2 = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \pi + k4\pi\\x = -\dfrac{\pi}{4} + k4\pi\\x = \dfrac{7\pi}{3} + k4\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\pi + k4\pi;\ -\dfrac{\pi}{4} + k4\pi;\ \dfrac{7\pi}{3} + k4\pi\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)