Recurrence Relation with cosine is it possible to find general solution of the recurrence relation such as
$a_{n}=a_{n-1}+A\centerdot \cos(a_{n-1})$
where $a_{0}=0$ and $A \ll 1$
EDIT:
At least for
$a_{n}=a_{n-1}+A - \frac{A}{2!}\centerdot a^2_{n-1} + \frac{A}{4!}\centerdot a^4_{n-1}$
where cosine is expanded with Taylor Series with 3 terms
 A: Since the comments above led to the concluson that good approximation(s) was something the OP was after, here are some.
Let $f:a\mapsto a+A\cdot\cos(a)$ and assume that $0\lt A\leqslant1$. Then $f$ is increasing on $[0,\pi/2]$ and $f(a)\gt a$ for every $a$ in $[0,\pi/2)$ hence the sequence $(a_n)$ defined by $a_0=0$ and $a_{n+1}=f(a_n)$ for every $n\geqslant0$ is increasing with limit $\pi/2$. 
Let $b_n=(\pi/2)-a_n$, then $b_0=\pi/2$, $(b_n)$ is decreasing to the limit $0$ and $b_{n+1}=b_n-A\cdot\sin(b_n)$.
Since $b_n=o(1)$, $\sin(b_n)=b_n+o(b_n)$ and $b_{n+1}=(1-A)\cdot b_n+o(b_n)$. Assume first that $\color{red}{0\lt A\lt1}$. Then a simple recursion yields
$$
\frac{b_n}{(1-A)^n}=B-\frac{A}{1-A}B_n,\quad B=\sum_{k=0}^{+\infty}\frac{b_k-\sin(b_k)}{(1-A)^k},\quad B_n=\sum_{k=n}^{+\infty}\frac{b_k-\sin(b_k)}{(1-A)^k}.
$$
Since $\sin(b)=b-\frac16b^3+o(b^3)$ when $b\to0$, one can show that $B$ converges and that $B_n$ is at most of order $(1-A)^{2n}$. Hence, with the finite $\color{red}{B\gt0}$ defined above,
$$
\color{red}{a_n=(\pi/2)-B\cdot(1-A)^n+O((1-A)^{3n})}.
$$
When $\color{blue}{A=1}$, $b_{n+1}=b_n-\sin(b_n)\sim\frac16b_n^3$ and one can show that $\log b_n=-c\cdot3^n+o(3^n)$ for some finite positive $c$, hence there exists some $\color{blue}{0\lt C\lt1}$ such that
$$
\color{blue}{a_n=(\pi/2)-C^{3^n+o(3^n)}}.
$$
