step by step solution:

the given equation is

x=3+\frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+\dots }}}}

let’s consider:

\textcolor{primary}{3+\frac{1}{2+\frac{1}{3+\dots }}=x}

after substituting the above value of **x** into the initial equation we get

x=\textcolor{primary}{3+\frac{1}{2+\frac{1}{{x} }}} \implies x=3+\frac{1}{\frac{2x+1}{x}}

x=3+\frac{1}{\frac{2x+1}{x}}\implies x=3+\frac{x}{2x+1}

x=3+\frac{x}{2x+1}\implies x=\frac{3(2x+1)+x}{2x+1}

x=\frac{3(2x+1)+x}{2x+1}\implies x(2x+1)=6x+3+x

x(2x+1)=6x+3+x\implies 2x^{2}+x=7x+3\implies 2x^2-6x-3=0

2x^2-6x-3=0

Solve the equation for **x** we get:

x=\frac{3-\sqrt{15}}{2}~~,~~\frac{3+\sqrt{15}}{2}

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