12. Find the domain of each of the following functions:
1) $y=\sqrt{1-x^{2}}$
2) $u=\frac{x-1}{x^{2}-5 x+6}+\sqrt[3]{2 x+1}$
3) $v=\arccos \frac{1-2 x}{3}$
4) $p=\frac{x}{\sin x}$
5) $q=\log _{2}\left(x^{2}-9\right)$.
Solution. 1) Since the argument $x$ is under a radical of even degree, the function $y$ will have real values only for those values of $x$ for which the radicand is non-negative, i.e., $1-x^{2} \geqslant 0$. Solving this inequality, we get

$$
x^{2} \leqslant 1 ; \quad|x| \leqslant 1 ; \quad-1 \leqslant x \leqslant 1
$$

Therefore, the domain of the function $y$ is the interval $[-1 ; 1]$.

2) Here the argument $x$ is in the denominator of the fraction. Therefore, $x$ cannot take those values that make the denominator zero, as division by zero is undefined. Setting the denominator to zero, we find these values of $x$:

$$
x^{2}-5 x+6=0 ; \quad x_{1}=2 ; \quad x_{2}=3
$$

The second term in the expression for the function $u$ does not impose any restrictions on the values of $x$, since the degree of the radical is odd. Therefore, the domain of the function $u$ is the entire number line, except for the points $x=2$ and $x=3$.

3) The function $v$ will be defined only for those values of $x$ for which $-1 \leqslant \frac{1-2 x}{3} \leqslant 1$. Solving these inequalities, we get

$$
-\frac{4}{3} \leqslant-\frac{2}{3} x \leqslant \frac{2}{3} ; \quad-1 \leqslant x \leqslant 2
$$

The interval $[-1 ; 2]$ is the domain of the function $v$.

4) Find the values of $x$ that make the denominator of the function $p$ zero: $\sin x=0 ; x_{k}=k \pi ; k=0, \pm 1, \pm 2, \ldots$

For these values of $x$, the function $p$ has no values.

The domain of the function $p$ is the entire number line, except for the points $x_{k}$.

5) The logarithmic function $q$ is defined only for positive values of its argument (the expression being logarithmed), so $x^{2}-9>0$.

Solving this inequality, we get $|x|>3$, from which it follows that $-\infty<x<-3$ and $3<x<+\infty$, i.e., the domain of the function $q$ consists of two infinite intervals $(-\infty ;-3)$ and $(3 ;+\infty)$.