1. The solution set of the equation $2|\sin x|^{1+\ln 2}=1+|\sin x|$ is $\qquad$
Since $|\sin x|<1$, we have $|\sin x|^{1+\ln 2}<|\sin x| \Rightarrow 2|\sin x|^{1+\ln 2}<2|\sin x|<1+|\sin x|$, thus $|\sin x|=1 \Rightarrow x=k \pi+\frac{\pi}{2}$, which means the solution set of the original equation is $\left\{x \left\lvert\, x=k \pi+\frac{\pi}{2}\right., k \in \mathbf{Z}\right\}$.