$A(\pi+x)=A(x)$ and $A(\frac{\pi}{2}-x)=A(x)$ implies ? $\forall x \in R:\ (\pi+x) \in ( \alpha+\pi Z) $ and $(\frac{\pi}{2}-x) \in ( \alpha+\pi Z )$ Here is exercise from french book 
Let $A(x)=\cos(x).\sin(x)$ for all $x\in\mathbb{R}$ knowing that :


*

*$$A\left(\dfrac{19\pi}{3}\right)=\dfrac{\sqrt{3}}{4}$$ 

*$$\forall x \in \mathbb{R}:\quad A(\dfrac{\pi}{2}-x)=A(x) \mbox{ and } A(\pi+x)=A(x). $$

*$$\forall x\notin \left(\dfrac{\pi}{2}+\pi\mathbb{Z} \right):\ A(x)=\dfrac{\tan(x)}{1+\tan^2(x)}$$


Question 


*

*$$\mbox{Solve}\quad  A(x)=\dfrac{\sqrt{3}}{4} \mbox{ over } ]-\pi;\pi]$$


Answer from the book:
since $A\left(\dfrac{19\pi}{3}\right)=\dfrac{\sqrt{3}}{4}$ then $$A(x)=\dfrac{\sqrt{3}}{4} \iff A(x)=A\left(\dfrac{19\pi}{3}\right)$$
from other hand $$\forall x \in \mathbb{R}:\quad A(\dfrac{\pi}{2}-x)=A(x) \mbox{ and } A(\pi+x)=A(x). $$ then 
$$ x=\dfrac{19\pi}{3}+k\pi \mbox{ or } x=\dfrac{\pi}{2}-\dfrac{19\pi}{3}+k\pi \mbox{ with } k\in\mathbb{Z}$$
My Question: 


*

*I can't understand that part of solution is that true in general if we have :


$$\forall x\in \mathbb{R}: \quad A(\pi+x)=A(x) \mbox{ and } A(\dfrac{\pi}{2}-x)=A(x)  \mbox{ and } A(x)=A(\alpha)$$ 
implies 
$$\forall x \in \mathbb{R}:\quad  \pi+x\in \left( \alpha+\pi \mathbb{Z}\right) \mbox{ and } \dfrac{\pi}{2}-x \in \left( \alpha+\pi\mathbb{Z}\right)$$
here is source of exercise: 

its solutions

 A: $$A(x)=\sin x \cdot \cos x= \frac 12 \cdot 2 \sin x \cdot \cos x=\frac 12 \sin 2x$$
$$A(x)=\frac{\sqrt{3}}{4}$$
$$\frac 12 \sin 2x=\frac{\sqrt{3}}{4}$$
$$\sin 2x=\frac{\sqrt{3}}{2}$$
$$2x=(-1)^k\arcsin \left(\frac{\sqrt{3}}{2} \right)+\pi k, k \in \mathbb Z$$
$$2x=(-1)^k \frac{\pi}{3} +\pi k, k \in \mathbb Z$$
$$x=(-1)^k \frac{\pi}{6} +\frac{\pi k}{2}, k \in \mathbb Z$$
$$x \in \left\{-\frac{5 \pi}{6}, -\frac{2 \pi}{3}, \frac{\pi}{6}, \frac{ \pi}{3} \right\}$$
Add
Let $T -$ period $f(x)=\sin 2x$. Then $T=\frac{2 \pi}{2}=\pi$. 
Then $\forall x \in \mathbb R$
$A(x)=A(x+\pi)$
$$A(\frac{\pi}{2}-x)=\sin 2(\frac{\pi}{2}-x)=\sin (\pi-2x)=\sin 2x =A(x)$$
If $x_0 \in (\alpha + \pi \mathbb Z)$, then $\pi +x_0 \in (\alpha + \pi \mathbb Z)$
Add2:
Let $x_0 -$ solution of $A(x)=\frac{\sqrt{3}}{4}$ and $\forall x \in \mathbb R$: $A(x)=A(x+\pi)$. Then $x_0+\pi -$ solution of $A(x)=\frac{\sqrt{3}}{4}$.
If $x_0 \in (\alpha + \pi \mathbb Z)$ then $x_0+\pi \in (\alpha + \pi \mathbb Z)$
Similarly for $x_0 \in (\alpha + \pi \mathbb Z)$ then $\frac{\pi}{2}-x_0 \in (\alpha + \pi \mathbb Z)$

