Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = x - \sin \left( {\pi + x} \right) + 2\cos \dfrac{{3\pi + x}}{2} - 3\\
= x - \sin \left[ {\pi - \left( {\pi + x} \right)} \right] - 2\cos \left( {2\pi + \dfrac{{x - \pi }}{2}} \right) - 3\\
= x - \sin \left( { - x} \right) - 2\cos \dfrac{{x - \pi }}{2} - 3\\
= x + \sin x - 2.\cos \left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right) - 3\\
= x + \sin x - 2\sin \dfrac{x}{2} - 3\\
y' = 1 + \cos x - 2.\left( {\dfrac{x}{2}} \right)'.\cos \dfrac{x}{2}\\
= 1 + \cos x - 2.\dfrac{1}{2}.\cos \dfrac{x}{2} = 1 + \cos x - \cos \dfrac{x}{2}\\
y' = 0 \Leftrightarrow 1 + \cos x - \cos \dfrac{x}{2} = 0\\
\Leftrightarrow 1 + \left( {2{{\cos }^2}\dfrac{x}{2} - 1} \right) - \cos \dfrac{x}{2} = 0\\
\Leftrightarrow 2{\cos ^2}\dfrac{x}{2} - \cos \dfrac{x}{2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos \dfrac{x}{2} = 0\\
\cos \dfrac{x}{2} = \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{2} = \dfrac{\pi }{2} + k\pi \\
\dfrac{x}{2} = \pm \dfrac{\pi }{3} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \pi + k2\pi \\
x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.
\end{array}\)
