1. In the set of real numbers, solve the equation:

$$
\frac{\sin x-\sqrt{\sin x}}{\cos x-\sqrt{\cos x}}=1
$$
1. For the left side to be defined, it must hold that $\sin x, \cos x \geqslant 0$ and $\cos x - \sqrt{\cos x} \neq 0$ (which reduces to $\cos x \neq 0$ and $\cos x \neq 1$). After setting these conditions, the equation reduces to $\sin x - \sqrt{\sin x} = \cos x - \sqrt{\cos x}$, or $\sin x - \cos x = \sqrt{\sin x} - \sqrt{\cos x}$, and finally

$$
(\sqrt{\sin x} - \sqrt{\cos x})(\sqrt{\sin x} + \sqrt{\cos x}) = \sqrt{\sin x} - \sqrt{\cos x}
$$

First, assume $\sqrt{\sin x} - \sqrt{\cos x} = 0$. From this, it follows that $\sin x = \cos x$, i.e., $x = \frac{\pi}{4} + k \pi, k \in \mathbb{Z}$, but due to the condition $\sin x, \cos x \geqslant 0$, only $x = \frac{\pi}{4} + 2 k \pi$, $k \in \mathbb{Z}$ remains.

Now assume $\sqrt{\sin x} - \sqrt{\cos x} \neq 0$. Then, by canceling out the common factor, the equation reduces to $\sqrt{\sin x} + \sqrt{\cos x} = 1$. Let $a = \sqrt{\sin x}$ and $b = \sqrt{\cos x}$. Then we have $a + b = 1$, where the numbers $a$ and $b$ satisfy $0 < a, b \leq 1$. Squaring both sides, we get $a^2 + b^2 + 2ab = 1$. Since $a^2 + b^2 = \sin x + \cos x \leq \sqrt{2}$, and $2ab = 2\sqrt{\sin x \cos x} \leq 1$, we have $a^2 + b^2 + 2ab \leq \sqrt{2} + 1$, which is a contradiction because $a^2 + b^2 + 2ab = 1$.

Therefore, the final solutions are only those obtained in the first case, i.e., $x = \frac{\pi}{4} + 2 k \pi$ for $k \in \mathbb{Z}$.