jmc

We have $$29\le x\lfloor x\rfloor <30.$$First, suppose that $x\ge 0.$ Then we have $x\lfloor x\rfloor \ge \lfloor x{\rfloor }^2,$ so $\lfloor x{\rfloor }^2<30,$ and $\lfloor x\rfloor \le 5.$ Also, $x\lfloor x\rfloor \le x^2,$ so $29\le x^2,$ which means that $\lfloor x\rfloor \ge 5.$ Thus, $\lfloor x\rfloor =5,$ so $\lfloor 5x\rfloor =29$ from the original equation. Thus, $29\le 5x<30,$ so $$5.8\le x<6.$$Indeed, if $5.8\le x<6,$ then $\lfloor x\lfloor x\rfloor \rfloor =\lfloor 5x\rfloor =29,$ so all $x\in [5.8,6)$ are solutions to the equation.

Now suppose that $x<0.$ Then we have $x\lfloor x\rfloor \le \lfloor x{\rfloor }^2,$ so $29\le \lfloor x{\rfloor }^2,$ and $\lfloor x\rfloor \le -6.$ But then $x<-5,$ so $$x\lfloor x\rfloor \ge -6x>-6(-5)=30,$$a contradiction. Hence, no negative $x$ satisfy the equation.

Thus, the solution set is the interval $\boxed{[5.8,6)}.$