$\sqrt{3x-2<x}$

Step 1 ：First, we need to ensure that both sides of the inequality are defined and non-negative. Since the square root is always non-negative, we need to find the domain of the square root function: $3x-2\ge 0$, which gives us $x\ge \frac{2}{3}$.

Step 2 ：Now, we need to find the domain of the inequality $x>3x-2$. Solving for x, we get $2x<2$, which gives us $x<1$.

Step 3 ：Taking the intersection of the domains, we get $\frac{2}{3}\le x<1$.

Step 4 ：Since both sides are non-negative (and defined) for $\frac{2}{3}\le x<1$, we can square both sides, to obtain the equivalent inequality $3x-2<x^2$.

Step 5 ：This simplifies to $x^2-3x+2>0$, which factors as $(x-1)(x-2)>0$.

Step 6 ：From the factored inequality, we can see that the inequality is satisfied when either both factors are positive ($x>2$) or both factors are negative ($x<1$).

Step 7 ：However, we must consider the domain we found earlier, $\frac{2}{3}\le x<1$. Taking the intersection with the inequality $x<1$, we get the final solution as $x\in \boxed{{\displaystyle [\frac{2}{3},1)}}$.