$$4\{x\}=[x]+\{x\}+[x]$$
$$3\{x\}=2[x]$$
$$\{x\}=\frac 23 [x]$$
$$0\le \frac 23 [x] <1$$
$$0\le [x]<1.5$$ So $[x]=0,1$
The solutions for $x$ should be infinite, but the given answer is 3.
Even if assume that the answer not being $\infty$ is upto interpretation, I still get only 2 as the answer. Please verify this solution.
The brackets represent fractional part and greatest integer part
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Using ${x}=x-[x]$, thee equation reduces to $$ [x]=\frac 35 x$$Now, $$x-1\le [x]\le x \\ \implies x-1\le \frac 35 x\le x \\ \implies 0\le x\le \frac 52$$We must also have $\frac 35x\in \mathbb Z$, and from here it is easy to deduce that $x=0$ and $x=\frac 53$ are the only solutions.
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Did you try to plot?
$$0\{x\}=x-[x]\to \\
4\{x\}=x+[x]\\
4(x-[x])=x+[x]\\4x-x=4[x]+[x]\\\frac{3}{5}x=[x]$$set $f(x)=\frac35 x$,$f(x)=[x]$plot them togheter ,you will see two solution, x=0 is trivial, (and) by considering the slope the other one is crosssection of $[x]=1,\frac35x=1$ so $x=\frac 53$
if you put $x=\frac 53\to \{x\}=\frac23, [x]=1$ you will have
$$4\times \frac 23=\frac53+1\checkmark$$
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I'll try to solve this from where you left.
You have rightly shown that:
$[x] =0,1$ whence you try to conclude that no. of solutions of the given equation is infinite (You thought so probably because $[x] =0\implies 0\le x\lt 1$.) That is wrong because you have only found $[x] $ and not $x$. In order to find $x$, you need $\{x\} $ also so that $x=[x] +\{x\} $.
Therefore, you actually want to find $x=y$ such that "$[y] =0$ and $4\{y\}=y+[y]$" or "$[y] =1$ and $4\{y\}=y+[y]$"
Case 1:$[y] =0$
$4\{y\}=y+[y]=y\implies 4\{y\}=\{y\}+[y]=\{y\}\implies \{y\} =0\implies y=\{y\} +[y] =0$
Case 2: $[y] =1$
$4\{y\}=y+[y] =y+1=\{y\}+2\implies 3\{y\}=2\implies y=\{y\}+[y]=2/3 +1=5/3$
Therefore, $\exists$only two solutions viz. $x=0$ and $x=5/3$.
Let $x=I+f$, with $I \in \Bbb{Z}$ and $0 \leq f <1$. Then we have $3f=2I$. Thus $f=0,\frac{2}{3}$ for $I=0,1$ respectively.
Thus only two solutions $0,1+\frac{2}{3}=\frac{5}{3}$.