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?

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

$$-\frac\pi2+2k\pi\le x\le\frac\pi2+2k\pi.$$