Answer
$\left\{\dfrac{2}{3}\right\}$
Work Step by Step
Squaring both sides, the given equation, $
2x=\sqrt{\dfrac{5x+2}{3}}
$, is equivalent to
\begin{align*}
(2x)^2&=\left(\sqrt{\dfrac{5x+2}{3}}\right)^2
\\\\
4x^2&=\dfrac{5x+2}{3}
.\end{align*}
In the form $ax^2+bx+c=0$, the equation above is equivalent to
\begin{align*}\require{cancel}
3\cdot4x^2&=\dfrac{5x+2}{\cancel3}\cdot\cancel3
\\\\
12x^2&=5x+2
\\
12x^2-5x-2&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(3x-2)(4x+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving the variable, then
\begin{array}{l|r}
3x-2=0 & 4x+1=0
\\
3x=2 & 4x=-1
\\\\
x=\dfrac{2}{3} & x=-\dfrac{1}{4}
.\end{array}
Since both sides of the original equation were raised to the second power, then checking of solutions is a must. Substituting the values of $x$ above in the original equation results to
\begin{array}{l|r}
\text{If }x=\dfrac{2}{3}: & \text{If }x=-\dfrac{1}{4}:
\\\\
2\left(\dfrac{2}{3}\right)\overset{?}=\sqrt{\dfrac{5\left(\dfrac{2}{3}\right)+2}{3}} & 2\left(-\dfrac{1}{4}\right)\overset{?}=\sqrt{\dfrac{5\left(-\dfrac{1}{4}\right)+2}{3}}
\\\\
\dfrac{4}{3}\overset{?}=\sqrt{\dfrac{\dfrac{10}{3}+\dfrac{6}{3}}{3}} & -\dfrac{1}{2}\overset{?}=\sqrt{\dfrac{\dfrac{5}{4}+2}{3}}
\\\\
\dfrac{4}{3}\overset{?}=\sqrt{\dfrac{\dfrac{16}{3}}{3}} & -\dfrac{1}{2}\ne\text{some nonnegative number.}
\\\\
\dfrac{4}{3}\overset{?}=\sqrt{\dfrac{16}{9}} &
\\\\
\dfrac{4}{3}\overset{\checkmark}=\dfrac{4}{3} &
.\end{array}
Since $x=-\dfrac{1}{4}$ DOES NOT satisfy the original equation then the only solultion is $x=\dfrac{2}{3}$. Hence, the solution set of the equation $
2x=\sqrt{\dfrac{5x+2}{3}}
$ is $\left\{\dfrac{2}{3}\right\}$.
