16. $\sin ^{2} 2^{\sqrt{-x}}=\frac{1}{2}$.
16. Assuming $2^{\sqrt{-x}}=y$, we get $\sin ^{2} y=\frac{1}{2}, y=\frac{\pi}{4}+\frac{\pi}{2} k$, $2^{\sqrt{-x}}=\frac{\pi}{4}+\frac{\pi}{2} k$. Since $2^{\sqrt{-x}} \geqslant 1$, the possible values for $k$ are: $k=1 ; 2 ; 3 ; \ldots, \sqrt{-x}=\log _{2}\left(\frac{\pi}{4}+\frac{\pi}{2} k\right) . \quad$ Hence, $x=-\log _{2}^{2}\left(\frac{\pi}{4}+\frac{\pi}{2} k\right)$.