solve the equation for x, where x is restricted to the given interval.\ny = sin(x + 1), for x in…

Explanation:

Step1: Apply inverse - sine function

We know that if $y = \sin(u)$, then $u=\sin^{-1}(y)+2k\pi$ or $u=\pi - \sin^{-1}(y)+2k\pi,k\in\mathbb{Z}$. Here $u = x + 1$, so $x+1=\sin^{-1}(y)+2k\pi$ or $x + 1=\pi-\sin^{-1}(y)+2k\pi$. But since $x\in\left[-\frac{\pi}{2}-1,\frac{\pi}{2}-1\right]$, we consider the principal - value of the inverse - sine function. So $x + 1=\sin^{-1}(y)$. $x+1=\sin^{-1}(y)$

Step2: Solve for x

Subtract 1 from both sides of the equation. $x=\sin^{-1}(y)-1$

Answer:

$x=\sin^{-1}(y)-1$