Why $P(z)\text{sin}(z)+Q(z)\text{cos}(z)$ has only finitely many non-real zeroes? Consider entire functions (defined and holomorphic on the whole complex plane) of the form:
$$f(z)=P(z)\text{sin}(z)+Q(z)\text{cos}(z),$$
where $P(z)$ and $Q(z)$ are polynomials with real coefficients. They seem to always have only finitely many zeros that don't lie on real axis. Namely, there is always some real $M>0$ so that if $|z|>M$ and $f(z)=0$, then $z\in\mathbb{R}$. How can I prove this?
I tried using Cauchy's integral theorem and integral formula, I tried inverting the function or using logarithms, but none worked. Intuitively, I think the point is that in order to split ("split" because the zeroes would obviously be symmetrical around the real axis) infinitely many zeroes that $\text{sin}$ and $\text{cos}$ have, you (in some way) have to introduce infinitely many saddles (zeros of the derivative), but polynomials can have only finitely many. Does this argument make sense and how could I turn it into a rigorous proof?
 A: I think that your conjecture is correct, and can be proved using Rouché's theorem.
Without loss of generality one can assume that $\deg Q < \deg P$, otherwise consider the function
$$
\begin{align}
 f(z +a) &= \left(  P(z) \cos(a) - Q(z) \sin(a) \right) \sin(z)
 + \left(  P(z) \sin(a) + Q(z) \cos(a) \right) \cos(z) \\
 &= \tilde P(z) \sin(z) + \tilde Q(z) \sin(z)
\end{align}
$$
with a suitably chosen $a \in \Bbb R$.
Then $f(z) = 0$ can be written as
$$ \tag{*}
 \tan(z) = r(z)
$$
where $r$ is a rational function with real coefficients and $\lim_{z \to \infty} r(z) = 0$. In particular we can find an $M > 0$ such that
$$ \tag{1}
|r(z)| < 1/2 \text{ for } |z| \ge M \, .
$$
We also need some estimates for the absolute value of the tangent. Using
$$
 \tan(x+iy) = \frac{\sin(2x) + i\sinh(2y)}{\cos(2x) + \cosh(2y)}
$$
from https://proofwiki.org/wiki/Tangent_of_Complex_Number we have
$$ \tag{2}
 |\tan(x+iy)| \ge \frac{|\sinh(2y)|}{\cosh(2y)+1} \text{ for } y \ne 0 \, ,
$$
and
$$ \tag{3}
 |\tan(x+iy)| = \frac{|\sinh(2y)|}{\cosh(2y)-1} > 1 \text{ for $x = \pi/2 + k \pi$ with $k \in \Bbb Z$.}
$$
Using $(2)$ and increasing $M$ if necessary, we have
$$ \tag{4}
 |\tan(x+iy)| > \frac 12 \text{for $|y| \ge M$.}
$$
It follows from $(1)$ and $(4)$ that $(*)$ has no solutions with $\operatorname{Im}(z) \ge M$.
Now consider the rectangles
$$
 R_k = \{ x+iy \mid \frac 12 \pi + k\pi \le x \le \frac 32 \pi + k\pi, |y| \le M \}
$$
with integers $k$ satisfying $\frac 12 \pi + k\pi > M$ or $\frac 32 \pi + k\pi < -M$.
Then $(1)$, $(3)$ and $(4)$ imply that $|\tan(z)| > 1/2 > r(z)$ for all $z$ on the boundary of $R_k$, and it follows from Rouchés theorem that $\tan(z) -r(z)$ and $\tan(z)$ have the same number of zeros in $R_k$, which is one.
On the other hand it follows from the intermediate value theorem that $\tan(x) - r(x)$ has one zero in the interval $\frac 12 \pi + k\pi \le x \le \frac 32 \pi + k\pi$. (This is where we use that the rational function is real on the real axis!)
So we have shown the following:

*

*$\tan(z) -r(z)$ has no zeros with $\operatorname{Im}(z) \ge M$.

*For sufficiently large $|k|$, $\tan(z) -r(z)$ has exactly one zero in each strip $\frac 12 \pi + k\pi \le x \le \frac 32 \pi + k\pi$, and that zero is real.

In particular, the equation $(*)$ has only finitely many non-real solutions.
