Can we prove that all zeros of entire function cos(x) are real from the Taylor series expansion of cos(x)?

Well, I think I can show it, but the idea requires knowing in advance that $\sin x$ has only real zeros:

We will use the following proposition ,which is given as an exercise in Ahlfors' text:

Show that if $f(z)$ is of genus $0$ or $1$ with real zeros, and if $f(z)$ is real for real z, then all zeros of $f'(z)$ are real. Hint: Consider $\text{ Im} \frac{f'(z)}{f(z)}$.

Integrating the Taylor series of the cosine gives the Taylor series of the sine:

$$\sin(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} z^{2n+1} $$

Since the coefficients are real, we see that the sine function is real for real arguments.

Using the formula $$\rho=\limsup_{n\to\infty}\frac{n\ln n}{-\ln|a_n|} $$ for the order of the entire function $\sum a_n z^n$, we can see that $\sin(z)$ has order $\rho=1$, and according to Hadamard's factorization theorem we find that its genus is $\leq 1$.

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In order to apply this on your example, you should ask whether $\xi(z)$ has an antiderivative with genus $\leq 1$, which vanishes exclusively on the real axis. (the real coefficients give the third condition automatically).

Hope this helps!

For $z=x+i y$, one verifies that $$ |\cos z|^{2}=\cos^{2} x+\sinh^{2} y,$$ by using the addition formula for cosine and Euler's formula (possible to deduce from the power series expansions of the involved functions). Consequently, one has the inequality $$|\cos z|^{2}\geq\sinh^{2} y.$$ Since $\sinh y=(e^{y}-e^{-y})/2=0$ iff $y=0$, the above inequality immediately implies that $\cos z$ cannot vanish for $\Im z\neq0$.