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.

The zeros of $\zeta(s)$ are located where the contours of $\Re(\zeta(s))=0$ intersect the contours of $\Im(\zeta(s))=0$ (see Contour Plots of the Zeta Function).

@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 $𝑘\pi$.