I'm self teaching and doing a book exercise which asks: "Considering only positive values of x, locate the first two turning points on the curve $2\sin x - x$ and determine whether they are maximum or minimum points"
My first step was to set the derivative to $0$ and solve:
$\frac{d}{dx}(2\sin x - x) = 2\cos x - 1$.
$2\cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2} \Rightarrow x = \frac{\pi}{3}$
To determine max/min I plugged $\frac{\pi}{3}$ into the second derivative:
$\frac{d}{dx}(2\cos x - 1) = -2\sin x$
$-2\sin \frac{\pi}{3} = -\sqrt{3}$ which is negative, so $\frac{\pi}{3}$ is a maximum point.
I then sketched $2\cos x - 1$ and figured the next turning point (i.e. when derivative crosses the $x$-axis) would be $2\pi - \frac{\pi}{3} = \frac{5}{3}\pi$ and figured as the previous point was a max, then this would be a min. (I felt like I was guessing a bit here and would have preferred to discover that analytically). My answers are therefore:
$\dfrac{\pi}{3}$ (max), $\dfrac{5\pi}{3}$ (min)
The back of the book gives these answers:
($\frac{\pi}{3}$, $\sqrt{3} - \frac{1}{3}\pi$), max
($\frac{5\pi}{3}$, $-\sqrt{3} - \frac{5}{3}\pi$), min
My question is where do these solutions that involve $\sqrt{3}$ come from? I did consider it to be a book error as it asks for first two turning points but there's 4 answers there.
