$\{x\}=x-[x]$. Then the equation $x+2\{x\}=3[x]$ has a solution of $\qquad$ .

Let's translate the solution part as well:

To solve the equation $x + 2\{x\} = 3[x]$, we start by expressing $x$ in terms of its integer and fractional parts:
\[ x = [x] + \{x\} \]

Substituting this into the equation, we get:
\[ [x] + \{x\} + 2\{x\} = 3[x] \]
\[ [x] + 3\{x\} = 3[x] \]
\[ 3\{x\} = 2[x] \]
\[ \{x\} = \frac{2}{3}[x] \]

Since $\{x\}$ is the fractional part of $x$, it must satisfy $0 \leq \{x\} < 1$. Therefore:
\[ 0 \leq \frac{2}{3}[x] < 1 \]
\[ 0 \leq [x] < \frac{3}{2} \]

Since $[x]$ is an integer, the only possible value for $[x]$ is 0 or 1.

1. If $[x] = 0$:
\[ \{x\} = \frac{2}{3} \cdot 0 = 0 \]
\[ x = [x] + \{x\} = 0 + 0 = 0 \]

2. If $[x] = 1$:
\[ \{x\} = \frac{2}{3} \cdot 1 = \frac{2}{3} \]
\[ x = [x] + \{x\} = 1 + \frac{2}{3} = \frac{5}{3} \]

Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad x = \frac{5}{3} \]

So, the solution to the equation is:
\[ \boxed{0 \text{ or } \frac{5}{3}} \]
3. $\frac{5}{3}$