Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$ 
Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$

Attempt:
$\star$ denotes $3\cdot 2^n+2\cdot 3^n$
$$\text{for  }n=1:\quad\star\equiv 5\not\equiv 1\\
\text{for  }n=2:\quad\star\equiv 5\not\equiv 1\\
\text{for  }n=3:\quad\star\equiv 3\not\equiv 1\\
\text{for  }n=4:\quad\star\equiv 0\not\equiv 1\\
\text{for  }n=5:\quad\star\equiv \color{red}1\\
\text{for  }n=6:\quad\star\equiv 5\not\equiv 1\\
\vdots$$
this is how should I aprroach this? or maybe there is smarter way?

You have the right idea. Working mod $7,\,$ by little Fermat $\,2^6\equiv 1\equiv 3^6\pmod 7\,$ so the sequence $\,f_n \equiv 3\cdot 2^n + 2\cdot 3^n \,$  has period $\,6,\,$ i.e $\, f_{n+6}\equiv f_n.\,$ So we need only check the first cycle for $\,n = 0,\ldots, 5.\,$ Using $\,2^3\equiv 1,\ 3^3\equiv -1\,$ we compute the cycle as
$$\begin{eqnarray} 3\ (1&,&2&,&4&,&\ \ \,1&,&\,\ \ 2&,&\,\ \ 4)\\
+\ 2\ (1&,&3&,&9&,&{-}1&,&{-}3&,&{-}9)\\
\equiv \ \  (5&,& 5&,& 2&,&\,\ \ \color{#c00}1&,&\,\ \ 0 &,&\,\ \ \color{#0a0}1)\end{eqnarray}$$
Therefore, we conclude that $\ f_n \equiv 1\iff $  $\,\color{#c00}{n\equiv 3}\ $ or $\,\ \color{#0a0}{n \equiv  5}\, \pmod 6$ 
