Answer
$\sin x=0$ or, $x=0, \pi, 2\pi,.....$
Work Step by Step
Given: $\sqrt {\sin^2 x+1}=\sqrt {\sin^2 x}+1$
The equation is not an identity.
$\sqrt {\sin^2 x+1}=\sqrt {\sin^2 x}+1$
This gives: $\sin^2 x+1=(\sin x+1)^2$
$\implies 2 \sin x=0$
This implies that
$\sin x=0$
Thus, $x=0, \pi, 2\pi,.....$
