This is going in the wrong direction.
Lemma: A function ##y(x)=\sin(\omega \cdot x +a)## has always the least period ##\dfrac{2\pi }{\omega}. ##
Proof:
\begin{align*}
y\left(x+n\cdot\dfrac{2\pi}{\omega}\right)&=\sin\left(\omega\left(x+n\cdot\dfrac{2\pi}{\omega}\right) + a\right)\\
&=\sin\left(\omega x+ 2\pi n +a \right)\\
&=\sin((\omega x+a)+2\pi n)\\
&=\sin(\omega x+a) \cos(2\pi n)+\cos(\omega x+a) \sin(2\pi n)\\
&=\sin(\omega x+a)\cdot 1 + \cos(\omega x+a) \cdot 0\\
&=\sin(\omega x+a)\\
&=y(x)
\end{align*}
Every function of period ##p## is also a function of period ##2p,3p,4p,\ldots## So we want to show now, that ##\dfrac{2\pi }{\omega}## is the least period. To do that, we first note that ##x= \dfrac{n\pi - a}{\omega}## are zeros of ##y(x)## for all integers ##n## so that the period could possibly be ##\dfrac{(n+1)\pi - a}{\omega}-\dfrac{n\pi - a}{\omega} = \dfrac{\pi}{\omega}.## We therefore calculate the slope at these points.
$$
y'(x)=\cos (\omega \cdot x +a)\cdot (\omega \cdot x +a)'= \omega \cos (\omega \cdot x +a)
$$
\begin{align*}
y'\left(\dfrac{-1\cdot\pi - a}{\omega}\right)&=\omega \cos \left( -\pi \right)=-\omega \\
y'\left(\dfrac{0\cdot\pi - a}{\omega}\right)&=\omega \cos \left( 0 \right) =\omega\\
y'\left(\dfrac{1\cdot\pi - a}{\omega}\right)&=\omega \cos \left( \pi \right) =-\omega
\end{align*}
This means, say that ##\omega > 0,## that ##y(x)## is always decreasing at the points ##x= \dfrac{(2n+1)\pi - a}{\omega}## and always increasing at ##x= \dfrac{2n\pi - a}{\omega}.## The period is thus at least
$$
\dfrac{(2n+1)\pi - a}{\omega}-\dfrac{(2n-1)\pi - a}{\omega}=\dfrac{2\pi}{\omega}
$$
##\square##
This means in our case with ##\omega=\dfrac{\pi}{2}## that we have a period of ##\dfrac{2\pi }{\omega}=\dfrac{2\pi }{\frac{\pi}{2}}=\dfrac{2}{\frac{1}{2}}=4.##
The period cannot be ##3.## I leave it up to you to find the exact point where your argument went wrong. The amplitude ##B(x)## must also be chosen with period ##4,## i.e. such that ##B(x)=B(x+4n).##
This is correct.
We must have ##B(11)=-2## and ##B(22)=-\dfrac{4\sqrt{3}}{3}=-\dfrac{4}{\sqrt{3}}.##
However, we also need the above condition that ##B(x)## is of period ##4##, i.e. that ##B(x)=B(x+4n)## for all ##n\in \mathbb{Z}.## You should be able by now to find such a periodic function ##B(x).##
What do you mean?
We defined ##z(x)=B(x)\cdot \sin\left(\dfrac{\pi\cdot x}{2}+\dfrac{\pi}{3}\right).## By setting ##B(x)=-2z(x) ## we would get
\begin{align*}
0&= B(x)\cdot \left(1+2\sin\left(\dfrac{\pi\cdot x}{2}+\dfrac{\pi}{3}\right)\right) \\
0&=B(22)\cdot \left(1+2\sin\left(\dfrac{\pi\cdot 22}{2}+\dfrac{\pi}{3}\right)\right)=-\dfrac{4}{\sqrt{3}}\cdot\left(1+2\cdot \left(-\dfrac{\sqrt{3}}{2}\right)\right)=-\dfrac{4}{\sqrt{3}}+4 \neq 0
\end{align*}
You defined ##B(x)## by using ##z(x)## but ##z(x)## is defined by ##B(x)## which is circular. And if ##B(x)\neq 0## we would have ##1=-2\sin\left(\dfrac{\pi\cdot x}{2}+\dfrac{\pi}{3}\right)## which can be solved for ##x.## We would be left with only a subset of all real numbers, but we want ##z(x)## to be defined everywhere.
You can use the sine term of the function ##z(x)## as a function of period ##4,## i.e. ##s(x):=\sin\left(\dfrac{\pi\cdot x}{2}+\dfrac{\pi}{3}\right).## However, ##s(11)=-\dfrac{1}{2}=\dfrac{1}{4}B(11)## and ##s(22)=-\dfrac{\sqrt{3}}{2}=-\dfrac{3}{2\sqrt{3}}=\dfrac{3}{8}B(22).## This means we have to adjust ##s(x).## I think the method in post #6 doesn't work in this case. So my first ansatz in the thread might be faster than shifting the wave to the right or to the top.
What we need is a function ##B(x)## with
$$
B(x)=B(x+4n)\text{ for all }n\in \mathbb{Z}\;\wedge\; B(11)=-2\;\wedge\; B(22)=-\dfrac{4}{\sqrt{3}}
$$
I would make an ansatz ##B(x)=p\cdot \sin\left(\dfrac{\pi\cdot x}{2}\right)+q\cdot \cos\left(\dfrac{\pi\cdot x}{2}\right).## This guarantees a period of ##4## by the Lemma, and we have two unknowns ##p,q## and two equations ##B(11)=-2## and ##B(22)=-\dfrac{4}{\sqrt{3}}.##
(I hope I made no serious typos, mistakes, or oversaw something. It's late here.)