Calculus - Find critical points I need help finding the critical points of this function: 
$f(x)=x-2 \sin x $
I found $f'(x)=1-2 \cos x $ and 
$f''(x)=2\sin x$
I know the next step is to set $f'(x)=0$ but when I do that I get $x=1.047$. 
But looking at the graph I see there is another critical point. How do I obtain this other point?
I also need to find the regions where the graph is concave upward and concave downward.
 A: So you want to solve
$$
0 = f'(x) = 1 - 2\cos(x)
$$
which means that 
$$
\cos(x) = \frac{1}{2}.
$$
This happens for example when $x = \frac{\pi}{3}\simeq 1.047$. But remember that $\cos$ is an even function so we also get the solution $x = -\frac{\pi}{3}$. Then remember that $\cos$ is $2\pi$ periodic, so we actually get infinitely many solutions:
$$
x = \pm\frac{\pi}{3} + n2\pi,
$$ 
where $n$ is any integer.
A: $$f'(x) = 1 - 2\cos x = 0 \implies \cos x = \dfrac 12 \implies x = \pi/3 \;\text{or}\; x  = 5\pi /3$$ for $x \in [0 , 2\pi]$, and in general $$x = \pm \frac \pi 3 + 2\pi n, \; n \in \mathbb Z$$
ADDED: Note that at $x = \pi/3, \;2\sin x = 2 \sin\left(\frac \pi 3\right) = \sqrt 3 > 0$, and at $x = \frac {5\pi}{3},\;\; 2\sin x = 2 \sin\left(\frac{5\pi}3\right) = -\sqrt 3 < 0$.  And in general $$x = \pi/3 + 2\pi n > 0 \implies f''(x) > 0,\quad x = -\pi/3 + 2\pi n \implies f''(x) < 0, \quad n\in \mathbb N$$
Now what does that tell you about when does $f(x) achieve its maximum, and when its minimum?
