First recall that [tex]\sin^{-1}x[/tex] is defined only for [tex]-1\le x\le1[/tex], which means [tex]\sin^{-1}(x+3)[/tex] requires that [tex]2\le x+3\le4[/tex].
Over this domain, [tex]\sin^{-1}(x+3)[/tex] will be invertible, and so taking the sine of both sides, we have
[tex]\sin^{-1}(x+3)=-\dfrac\pi6\implies\sin\left(\sin^{-1}(x+3)\right)=\sin\left(-\dfrac\pi6\right)[/tex]
[tex]x+3=-\sin\dfrac\pi6[/tex]
[tex]x+3=-\dfrac12[/tex]
[tex]x=-\dfrac72[/tex]
