Find domain of $ y = \arcsin\big(\frac{2}{2+\cos x}\big)$ what is a principle of fining domain of Trigonometric func or inverse (arcsin,...)? I need find domain of this funciton
$$ y = \arcsin\left (\frac{2}{2+\cos x}\right)$$
First step i do this: $$\arcsin(x) \in [-1,1]$$
$$-1 \leq \frac{2}{2+\cos x} \leq 1$$ =>
$$\cos x \geq 4 $$ and $$\cos x \geq 0$$
Am i true?
 A: You are almost correct. There was a sign error when you applied the left inequality.
$-1\le\sin(\theta)\le1$. This says that $\arcsin(x)$ can only be defined for $-1\le x\le1$. So $\arcsin\left(\frac2{2+\cos(x)}\right)$ to be defined, we need
$$
-1\le\frac2{2+\cos(x)}\le1\\
\Updownarrow\\
-2-\cos(x)\le2\le2+\cos(x)\\
$$
The left inequality says that $\cos(x)\ge-4$ (which is always true).
The right inequality says that $\cos(x)\ge0$.
Thus the domain of $\arcsin\left(\frac2{2+\cos(x)}\right)$ would be the set where $\cos(x)\ge0$.
As you comment, the domain is therefore,
$$
\bigcup_{k\in\mathbb{Z}}\left[2k\pi-\frac\pi2,2k\pi+\frac\pi2\right]
$$
A: To be systematic, you need to evaluate the domain in all steps of the computation, starting from the "innermost" expressions.
We have a transcendental function ($\arcsin$) of a quotient where the denominator is a constant plus a transcendental function ($\cos$) of the variable.
The domain of the cosine is the whole of $\mathbb R$.
So is the domain of the sum $2+\cos(x)$.
The domain of the quotient is any value of the numerator and denominator, except where the denominator is zero, $2+\cos(x)=0$. This never happens, as $\cos(x)\ge-1$.
And finally, the domain of the argument of the arc sine is $[-1,1]$, which requires
$$-1\le\frac2{2+\cos(x)}\le1.$$
As the denominator is positive, we  can rewrite
$$-2-\cos(x)\le2\le2+\cos(x),$$ or
$$-4\le\cos(x)\land0\le\cos(x).$$
The first inequality always holds, while the second is achieved when
$$-\frac\pi2+2k\pi\le x\le\frac\pi2+2k\pi.$$
