#### Answer

$x= \dfrac{(2n+1)\pi}{2}$
Where $n$ is an integer.

#### Work Step by Step

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.