11. If $[x]$ represents the integer part of the real number $x$, and $\{x\}=x-[x]$ represents the fractional part of the real number $x$, determine the solution set of the equation $[x]\{x\}+x=2\{x\}+6$.

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Solve
$$
[x]\{x\}+x=[x]\{x\}+[x]+\{x\} \Rightarrow([x]-1)(\{x\}+1)=5 \Rightarrow[x]=4,5,6 \text {. }
$$

On the other hand, $[x]\{x\}+x=[x](x-[x])+x=2(x-[x])+6 \Rightarrow[x]^2-(x+2)[x]+x+6=0$, when $[x]=4$, $16-4(x+2)+x+6=0 \Rightarrow x=\frac{14}{3}$; when $[x]=5$, $25-5(x+2)+x+6=0 \Rightarrow x=\frac{21}{4}$; when $[x]=6$, $36-6(x+2)+x+6=0 \Rightarrow x=6$.
In summary, the solution set of the equation $[x]\{x\}+x=2\{x\}+6$ is $\left\{\frac{14}{3}, \frac{21}{4}, 6\right\}$.