1. If $2|\sin x|^{1+182}=1+|\sin x|$, then the solution set of $x$ is
1. $\left\{x \left\lvert\, x=\frac{\pi}{2}+k \pi\right., k \in Z\right\}$.

Since $|\sin x|^{1 / 2} \leqslant 1$, then $2|\sin x|^{1-182} \leqslant 2|\sin x| \leqslant 1+\sin |x|$.
Therefore, the left side of the equation equals the right side if and only if $|\sin x|=1$, at which point the solution set is $\left\{x \left\lvert\, x=\frac{\pi}{2}+k \pi\right.\right.$, $k \in Z\}$.