$\sqrt{3 x-2< x}$
Answer
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 \(x \in \boxed{\left[\frac{2}{3}, 1\right)}\).
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 \(x \in \boxed{\left[\frac{2}{3}, 1\right)}\).
