Question

$x \in \left[-\frac{2029\pi}{2}; -\frac{2027\pi}{2}\right]$

Answer 1

x+1014\pi

Answer 2

For any $x \in [n\pi-\frac{\pi}{2}, n\pi+\frac{\pi}{2}]$, we have:

$$\sin^{-1}(\sin x)=(-1)^n (x-n\pi).$$

For the interval

$$x\in\Big[-\frac{2029\pi}{2}, -\frac{2027\pi}{2}\Big],$$

set

$$n\pi-\frac{\pi}{2}=-\frac{2029\pi}{2}\quad\Longrightarrow\quad n\pi=-\frac{2029\pi}{2}+\frac{\pi}{2} = -\frac{2028\pi}{2}=-1014\pi.$$

Thus, $n=-1014$ and consequently,

$$n\pi+\frac{\pi}{2}=-1014\pi+\frac{\pi}{2}=-\frac{2027\pi}{2}.$$

Since $-1014$ is even, $(-1)^n = (-1)^{-1014}=1$. Therefore:

$$\sin^{-1}(\sin x)= x-(-1014\pi)= x+1014\pi.$$