Let us now consider the problem (\ref{eqn1-1}) and the series:
\begin{eqnarray}\label{eqn5-1}
u(x, t)=\sum\limits_{n=1}^{\infty}\left\{f_n\cos(nt)+\frac{h_n-f_n\cos n}{\sin n}\sin( nt)\right\}\sin(nx),
\end{eqnarray}
where $f_n$ are given by the formula (\ref{eqn4-2}), $\displaystyle h_n=\frac{2}{\pi}\int\limits_0^{\pi}h(x)\sin(nx)dx. $

Note that the series (\ref{eqn5-1}) upon formal substitution into all relations (\ref{eqn1-1}) turns them into true equality.

We have not defined the classes of functions $f(x), h(x)$ for solving the problem (\ref{eqn1-1}) yet. We turn to the solution of the problem (\ref{eqn4-1}): $f(x)\in H_0^4[0;\pi],$ $g(x)\in H_0^3[0;\pi]$ to determine them.  However, the series (\ref{eqn5-1}) does not contain components of the expansion of the function $g(x)$ into a Fourier series. We compare the series (\ref{eqn5-1}) and the series (6) in order to determine the components $g_n:$
\begin{eqnarray}\label{eqnGn}
\sum\limits_{n=1}^{\infty}\left\{f_n\cos(nt)+\frac{h_n-f_n\cos n}{\sin n}\sin(nt)\right\}\sin(nx)=\\=\sum\limits_{n=1}^{\infty}\left\{f_n\cos(nt)+\frac{g_n}{n}\sin(nt)\right\}\sin(nx).
\end{eqnarray}
If we choose
\begin{eqnarray}\label{eqnGnF}
g_n=\frac{h_n-f_n\cos(n)}{\sin n}\cdot n,
\end{eqnarray}
then the solutions of the Dirichlet (\ref{eqn1-1}) and Cauchy (\ref{eqn4-1}) problems will coincide. Let us find out under what conditions this is possible.

We will use a result from number theory for what follows:
\begin{theorem}\label{Dirichlet}
There are only finitely many non-multiplicative numbers $n, m\in\mathbb{N},$ such that $\displaystyle\Big|\pi-\frac{m}{n}\Big|\leq\frac{C_{\pi}}{n^{\mu}},$ $2 \leq \mu \leq 7.7.$
\end{theorem}
\begin{proof}[Proof:]
According to a result of \cite{Japan}, there is a number $C_{\pi}>0,$ such that there are only finitely many non-multiplicative numbers $n, m\in\mathbb{N},$ such that $\displaystyle\Big|\pi-\frac{m}{n}\Big|\leq\frac{C_{\pi}}{n^{7.7}}.$ Therefore, $\exists N,$ such that $\forall n>N$ $\displaystyle\Big|\pi-\frac{m}{n}\Big|\geq\frac{C_{\pi}}{n^{\mu}}.$

Note also that the number $\pi$ is transcendental, and the degree of its irrationality $\mu$ cannot be less than 2. Thus, $\mu\geq 2.$

From these two statements we obtain that $\mu\in [2; 7.7].$
\end{proof} 
\begin{lemma}\label{h10} Let the function $\displaystyle \eta(x)\equiv\sum\limits_{n=1}^{\infty}\frac {h_n-f_n\cos n}{\sin n}\sin(nx),$ $h(x), f(x) \in H_0^k[0; \pi],$ where $k=\mu+3,$ if $\mu$ is an integer, and $k=[\mu]+4,$ if it is not an integer. Then $\eta(x)\in H^3_0[0; \pi].$
\end{lemma} 
\begin{proof}[Proof:]
Consider the series
\begin{eqnarray}\label{eqn5-6} 
\eta(x)=\sum\limits_{n=1}^{\infty}\frac{h_n-f_n\cos n}{\sin n}\ sin(nx).
\end{eqnarray}
Since $\exists N,$ that $\forall n, m>N$ $\displaystyle\Big|\pi-\frac{m}{n}\Big|\geq\frac{C_{ \pi}}{n^{\mu}},$ then $\displaystyle |\pi n- m|\geq\frac{C_{\pi}}{n^{\mu-1}}.$ Let's choose the numbers $n$ such that the difference $\displaystyle 0<|\pi n - m |<\frac{\pi}{2}.$ The function $\sin(a)$ is increasing provided that $\displaystyle 0< a< \frac{\pi}{2}$, so the following is true:
\begin{eqnarray*}
\sin(|\pi n-m|)\geq \sin\left(\frac{C_{\pi}}{n^{\mu-1}}\right), 
\end{eqnarray*}
whence for sufficiently large $n:$
\begin{eqnarray*}
|\sin(m)|\geq \sin\left(\frac{C_{\pi}}{n^{\mu-1}}\right).
\end{eqnarray*}
Therefore,
\begin{eqnarray}\label{eqn5-4}
\frac{1}{|\sin(m)|}\leq 2\frac{n^{\mu-1}}{C_{\pi}}.
\end{eqnarray}
Note that the sequence of numbers $n$ is determined by the value $m,$ but in order for the condition $|\pi n-m|\leq\pi/2$ to be satisfied, $n$ will not exceed $m.$
Thus, from this reasoning and (\ref{eqn5-4}) for numbers $n$ starting from some value $N$ the following relation is true:
\begin{eqnarray}\label{eqn5-4a}
\frac{1}{|\sin(n)|}\leq 2\frac{n^{\mu-1}}{C_{\pi}}.
\end{eqnarray}
The estimate $\forall n > N$ follows from (\ref{eqn5-4a}) and (\ref{eqnGnF}) :
\begin{eqnarray}\label{eqn5-5}
\Big|\frac{h_n-f_n\cos n}{\sin n}\Big|\leq\frac{2n^{\mu-1}}{C_{\pi}}(h_n-f_n\cos n).
\end{eqnarray}
Since only the estimate for $\mu$ is known, we choose $f(x), h(x)\in H^{12}_0[0; \pi].$ This estimate may be greatly overestimated, but we have no other data.

Then the relation is true: 
\begin{eqnarray*} 
\Phi_n=\frac{2}{\pi}\int\limits_0^{\pi}f^{(12)}(x)\sin(nx)dx, \Psi_n =\frac{2}{\pi}\int\limits_0^{\pi}h^{(12)}(x)\sin(nx)dx, \\ \sum\limits_{n=1}^{\ infty}\Phi_n^2<C_1, \sum\limits_{n=1}^{\infty}\Psi_n^2<C_1, f_n=\frac{\Phi_n}{n^{12}}, h_n=\frac {\Psi}{n^{12}}.
\end{eqnarray*}
And the chain of estimates is correct: 
\begin{eqnarray*} \eta'''(x)=\sum\limits_{n=1}^{\infty}\frac{h_n-f_n\cos(n )}{\sin n}(-n^3)\cos(nx)\cdot n=\\=\sum\limits_{n=1}^{\infty}\frac{\Psi_n-\Phi_n\cos(n)} {n^{12}\sin n}(-n^3)\cos(nx),\\ \big|\sum\limits_{n=1}^{\infty}\frac{\Psi_n-\Phi_n\ cos(n)}{n^{12}\sin n}(-n^3)\cos(nx)\big|\leq\\
\leq \sum\limits_{n=1}^{\infty}\frac{|\Psi_n-\Phi_n\cos(n)|n^{ 8}}{n^{12}}\frac{(n^3)}{C_{\pi}}\leq\sum\limits_{n=1}^{\infty}\frac{|\Psi_n-\ Phi_n\cos(n)|}{n\cdot C_{\pi}}.
\end{eqnarray*}
This series is convergent, therefore, $\eta(x)\in H_0^3[0; \pi].$
\end{proof}
\begin{lemma}\label{Existance}
Let $f(x), h(x)\in H^{12}_0[0; \pi].$ Then the solution (\ref{eqn1-1}) exists.
\end{lemma}
\begin{proof}[Proof:]
By Lemma \ref{h10}, if $f(x), h(x)\in H_0^{12} [0; \pi],$ then $\exists g(x)\in H^3_0[0; \pi].$ By Lemma 2, the solution (\ref{eqn4-1}) exists and is unique in this case. Moreover, the constructed solution satisfies all the conditions of the problem (\ref{eqn1-1}). Therefore, it is also a solution to the problem (\ref{eqn1-1}). Thus, a solution to  (\ref{eqn1-1}) exists.
\end{proof}