Find the coordinates of the inflexion points of $A(\beta)=8\pi-16\sin(2\beta)$ in $\mathbb{R}$

Question

I tried:

$$A'(\beta) = -32\cos(2\beta)$$ $$A''(\beta) = (-32\cos(\beta))' = 64\sin(2\beta)$$ $$\\$$ $$\\0 = \sin(2\beta) \Leftrightarrow \\ 2\beta = \arcsin(0) +2k\pi \lor 2\beta = \pi - \arcsin(0) +2k\pi \Leftrightarrow \\ \beta = k\pi \lor \beta = \frac{\pi}{2}+k\pi $$

$k \in \mathbb{Z}$

So the coordinates would be $$(a,8\pi), a = k\pi \lor a = \frac{\pi}{2}+k\pi$$

But my book says the solution is $$(a,8\pi), a = k\frac{\pi}{2},k\in\mathbb{Z}$$

What did I do wrong?

Answer 1

$A''(\beta) = 64\sin(2\beta) = 0$

Let $\theta = 2\beta$, then where is $\sin\theta = 0$?

$\theta = k\pi$, where $k \in \mathbb{Z}.$ Thus $\beta = k\frac{\pi}{2}$