## Calculus (3rd Edition)

$x= \dfrac{(2n+1)\pi}{2}$ Where $n$ is an integer.
Given $$y=x-2 \cos x$$ Since \begin{align*} y'&=1+2\sin x \\ y''&=2\cos x \end{align*} Then $f''(x)=0$ for $x=\dfrac{(2n+1)\pi}{2}$ and $$y''>0\ \ \text{ for } \ x\in \left(\frac{(4n-1)\pi}{2},\frac{(4n+1)\pi}{2}\right) \\ y''<0\ \ \text{ for } \ x\in \left(\frac{(4n+1)\pi}{2},\frac{(4n+3)\pi}{2}\right)$$ Hence there exists an inflection point at $x= \dfrac{(2n+1)\pi}{2}$ Where $n$ is an integer.