I'm struggling to understand why my approach doesn't produce the exact answer for the following equation:
(The brackets "[.]" denote the floor function and "{.}" denotes the fractional part function.)
$3[x]-2x=\{x\}+1/2$
My approach was to rewrite the equation as:
$6[x]-4x=2\{x\}+1$
Then, by substituting $[x]=x-\{x\}$, I arrived at:
$\frac{2x-1}{8}=\{x\}$
Given that $0≤\{x\}<1$,
$=>0≤\frac{2x-1}{8}<1$
$=>1/2≤{x}<9/2$
It turns out that even though the accurate solution set, i.e., $\{7/6,5/2,23/6\}$, falls within the above interval, this procedure doesn't directly yield those values. What might be the error?
You haven't made an error. You've just shown that "if $3[x]−2x=\{x\}+\frac{1}{2}$, then $\frac{1}{2}\leq x<\frac{9}{2}$". But this is need not be an "if and only if", so it need not yield the precise solution set you desire. At this stage, your solution is simply incomplete rather than incorrect.
To illustrate this with an extremely silly example, suppose I want to solve $x^{2} - 1 = 0$ for $x \in \mathbb{R}$. Observe the following reasoning:
All the reasoning is correct. Any solution to $x^{2} - 1 = 0$ must necessarily not be equal to $0$. However, the condition $x \neq 0$ is not sufficient to solve the equation. I recommend giving Wikipedia's Necessity and sufficiency or Wikipedia's If and only if a read if you have the time.
Let us now finish up the solution to your problem. You are given that $x\in\mathbb{R}$ satisfies $$ 3[x] - 2x = \{x\} + \frac{1}{2}, \qquad (1) $$ and you correctly deduced that such an $x$ must necessarily satisfy $\frac{1}{2} \leq x < \frac{9}{2}$.
Use again the fact that $[x] = x - \{x\}$ on Equation $(1)$ to obtain $$ \frac{3}{4}x + \frac{1}{8} = [x]. \qquad (2) $$ Now since we know that $\frac{1}{2} \leq x < \frac{9}{2}$, Equation $(2)$ gives us $$ \frac{3}{4}\left(\frac{1}{2}\right) + \frac{1}{8} \leq [x] < \frac{3}{4}\left(\frac{9}{2}\right) + \frac{1}{8}, $$ which simplifies to $$ \frac{1}{2} \leq [x] < 3 + \frac{1}{2}. $$ In particular, because $[x]$ must be an integer, we require $$ [x] = 1 \quad \text{or}\quad [x]=2 \quad \text{or}\quad [x] = 3. $$ If $[x] = 1$, then Equation $(2)$ yields $x = \frac{7}{6}$. On the other hand, if $[x] = 2$, then Equation $(2)$ yields $x = \frac{5}{2}$. Finally, if $[x] = 3$, then Equation $(2)$ yields $x = \frac{23}{6}$. At this stage, we can conclude that we must necessarily have $$ x = \frac{7}{6} \quad \text{or} \quad x = \frac{5}{2} \quad \text{or} \quad x = \frac{23}{6}. $$ We then check that all these are solutions to Equation $(1)$. Indeed, we have $$ 3\left[\frac{7}{6}\right] - 2\left(\frac{7}{6}\right) = \frac{2}{3} = \left\{\frac{7}{6}\right\} + \frac{1}{2}, $$ and $$ 3\left[\frac{5}{2}\right] - 2\left(\frac{5}{2}\right) = 1 = \left\{\frac{5}{2}\right\} + \frac{1}{2}, $$ and $$ 3\left[\frac{23}{6}\right] - 2\left(\frac{23}{6}\right) = \frac{4}{3} = \left\{\frac{23}{6}\right\} + \frac{1}{2}. $$
Therefore, for any $x \in \mathbb{R}$, $$ 3[x] - 2x = \{x\} + \frac{1}{2} \quad \text{if and only if}\quad x \in \left\{\frac{7}{6}, \frac{5}{2}, \frac{23}{6}\right\}. $$ This gives $\left\{\frac{7}{6}, \frac{5}{2}, \frac{23}{6}\right\}$ as our precise solution set to Equation $(1)$.