Let $k\leq 2x < k+1\iff \frac{k}{2}\leq x < \frac{k+1}{2}$ for some integer $k$. We do casework on if $k$ is odd or even. Case: $k$ is odd We have that $\frac{k-1}{2} + k = x\iff x = \frac{3k-1}{2}$. Note that $k\leq 3k-1\iff 1\leq k$. Also, $3k-1 < k+1\iff 2k < 2 \iff k \leq 0$. These two conditions clearly cannot be met at the same time and so there's no solution for this case. Case: $k$ is even We have that $\frac{k}{2}+k=x\iff x = \frac{3k}{2}$. Note that $k\leq 3k\iff 0\leq k$. Also, $3k < k+1\iff 2k < 1 \iff k \leq 0$. The only solution for $k$ here is $0$. Then, $x=\frac{3\cdot 0}{2}=0$. The only solution to the equation is $\boxed{x=0}$.