You can try to motivate your statement or present evidence that supports it, in order to get people to understand better what you are asking.

@Zima Thank you for your comment. As I wished to better understand why x in non trivial zeros is always 1/2 , I expressed the imaginary part of the exponent in the goniometric form. Then I separated the sine part and the cosine part as they should be both zero. Then I expressed the cosine function using the sine function. By playing with wolfram, I noticed that fi can be not only pi/2 but any angle different from an integer multiple of pi. So I would like to check whether this behaviour is well known.

"As I wished to better understand why x in non trivial zeros is always 1/2" We do not know this. That is the famous unproven Riemann Hypothesis.

@Gary Thank you for your comment. I am aware of that. I mean that all non trivial zeros of the Riemann zeta function known so far have x = 1/2.

Do you mean is it correct to argue the following assuming the Riemann hypothesis?

@StevenClark I believe that it is independent.

I'm not sure I understand what you're trying to do, but its true the Dirichlet eta function $$\eta(s)=\left(1-2^{1-s}\right) \zeta (s)=\sum\limits_{n=1}^{\infty} (-1)^{n+1}\, n^{-s}\,,\quad\Re(s)>0$$ has the same zeros as the Riemann zeta function in the critical strip $0<\Re(s)<1$, and it's also true that $$n^{-s}=n^{-\Re(s)}\, (\cos(-\Im(s)\, \log(n))+i\, \sin(-\Im(s)\, \log(n)))$$ so the Dirichlet eta function can also be evaluated as $$\eta(s)=\sum\limits_{n=1}^{\infty} (-1)^{n+1}\, n^{-\Re(s)}\, (\cos(-\Im(s)\, \log(n))+i\, \sin(-\Im(s) \log (n))),\quad\Re(s)>0.$$

@StevenClark Many thanks for your summary. I will explain my approach in the question's section.

It doesn't really matter if you can evaluate $\sum\limits_{n=1}^{\infty} (-1)^{n+1}\, n^{-s}$ first and then separate the real and imaginary parts at the end, or evaluate the real and imaginary parts separately along the way. You can also apply trigonometric identities to make $\eta(s)=\sum\limits_{n=1}^{\infty} (-1)^{n+1}\, n^{-\Re(s)}\, (\cos(-\Im(s)\, \log(n))+i\, \sin(-\Im(s) \log (n)))$ look like something different, but I don't really see the point.

@StevenClark Thank you for your comments. I agree with both of them. However, my question is about something a little bit different. I have explained the matter in the question's section in detail. Could you please look at the updated question's section again?

I don't understand your motivation or what you're doing. You can't shift the sine functions phases by an integer multiple $k$ of $\frac{\pi}{\ln(n)}$ because for $n=1$ you end up with the infinite expression $\frac{k\, \pi}{0}$.

@StevenClark Thank you for your comment. I can avoid the infinite expression by starting with $n=2$ because the first term is $0$ anyway or by leaving factoring out the term $\ln(n)$ in the argument of the sine function and working with $\pi$ or $\varphi$ directly (which I have done). My motivation is a better understanding why $x$ of all known non trivial zeros is $\frac{1}{2}$. The conclusion is that shifting the sine fuction retains the non trivial zeros. The sine function and the cosine function in the Dirichlet eta function are shifted by $\frac{\pi}{2}$ only.

You need to account for the $n=1$ term which can be accomplished as follows $$\eta (s)=1-\sum\limits_{n=2}^{\infty}\, (-1)^n\, n^{-\Re(s)} \left(\sin\left(\frac{\pi}{2}-\Im(s) \log(n)\right)+i\, \sin(-\Im(s) \log(n))\right)$$ but I don't understand what you're doing with the phase shifting.

For $$f(s)=1-\sum\limits_{n=2}^{\infty}\, (-1)^n\, n^{-\Re(s)} \left(\sin\left(\frac{\pi}{2}-\Im(s) \log(n)+\varphi_1\right)+i\, \sin(-\Im(s) \log(n)+\varphi_2)\right)$$ can you define examples of the values of the pair $\left(\varphi_1,\varphi_2\right)$ that you think both preserve and don't preserve the locations of the zeta-zeros? The only thing obvious to me is $\left(\varphi_1,\varphi_2\right)=\left(2 k_1 \pi,2 k_2 \pi\right)$ where $k_1,k_2\in\mathbb{Z}$ preserves the locations of the zeta-zeros since in this case $f(s)$ evaluates to $\eta(s)$.

However, any shift of the sine function, except for an interger multiple of $\pi$, results in the same non trivial zeros as the shift by $\frac{\pi}{2}$. This means that $𝑥=\frac{1}{2}$ holds more generally.

I use separate equations: $$\sum_{n=1}^{\infty}(-1)^{n}n^{-x}\sin(y\ln(n))=0,\\ $$ $$ \sum_{n=1}^{\infty}(-1)^{n}n^{-x}\sin(y\ln(n)+\varphi)=0.\\ $$ If $\varphi\neq k\pi$, non trivial zeros are retained. If you use $\varphi_1$ and $\varphi_2$, $\varphi_1-\varphi_2$ should differ from $ k\pi$.

I have found that a phase shift is defined as $\frac{\varphi}{ln(n)}$ in $\sin(\ln(n)(x+\frac{\varphi}{ln(n)}))$. However, I believe that I can call $\varphi$ a phase shift as well (at least informally).

I have just tested it for $\varphi=0.001\pi$ and non trivial zeros are still there.

Now I have noticed that you use $\frac{\pi}{2}$ in your equation as well. Then $\frac{\pi}{2}+ \varphi_1−\varphi_2$ should differ from $𝑘\pi$.