Let us consider the first equation:
$$a \sin x + b \cos x + c = 0.$$
This can be rewritten as:
$$a \sin x + b \cos x = -c.$$
Let $R = \sqrt{a^2 + b^2}$ and let $\varphi$ be such that $a = R \sin \varphi$, $b = R \cos \varphi$.
Then:
$$a \sin x + b \cos x = R \sin x \sin \varphi + R \cos x \cos \varphi = R \cos(x - \varphi).$$
So the equation becomes:
$$R \cos(x - \varphi) = -c.$$
This equation has a real solution for $x$ if and only if $|c| \leq R$.

Suppose $|c| > R$. Then the first equation has no real solution.

Now consider the second equation:
$$\text{a y + b y + 2c = 0.}$$
Let $t = \tan y$, $t \in \mathbb{R}$, $t \neq 0$.
Then $\text{y = t}$, $\ctg y = 1/t$.
So the equation becomes:
$$a t + \frac{b}{t} + 2c = 0.$$
Multiply both sides by $t$ (for $t \neq 0$):
$$a t^2 + 2c t + b = 0.$$
This is a quadratic equation in $t$ with discriminant:
$$\Delta = (2c)^2 - 4a b = 4c^2 - 4a b = 4(c^2 - a b).$$
Since $a \neq 0$, this quadratic equation has real solutions if and only if $c^2 - a b \geq 0$.

But recall that we are considering the case $|c| > R = \sqrt{a^2 + b^2}$.
So $c^2 > a^2 + b^2$.
Therefore,
$$c^2 - a b > a^2 + b^2 - a b = (a - b)^2 \geq 0.$$
So the discriminant is positive, and the quadratic equation has real solutions for $t \neq 0$.

Therefore, at least one of the two equations has real solutions.