Question

if sin^-1 (x+3) = -pi/6 , x = ?

Answer 1

First recall that $\sin^{-1}x$ is defined only for $-1\le x\le1$, which means $\sin^{-1}(x+3)$ requires that $2\le x+3\le4$.

Over this domain, $\sin^{-1}(x+3)$ will be invertible, and so taking the sine of both sides, we have

$\sin^{-1}(x+3)=-\dfrac\pi6\implies\sin\left(\sin^{-1}(x+3)\right)=\sin\left(-\dfrac\pi6\right)$
$x+3=-\sin\dfrac\pi6$
$x+3=-\dfrac12$
$x=-\dfrac72$