$ f(x)$ is  a continuous function with the piriodicity of $ 2\pi$ and $ c$ is a positive constant number.
Find $ f(x)$ and $ c$ such that $ \int_0^{2\pi} f(t \minus{} x)\sin tdt \equal{} cf(x)$ with $ f(0) \equal{} 1$ for all real numbers $ x$.
1. **Expressing \( f(x) \) and \(\sin(t)\) as Fourier series:**

   Given that \( f(x) \) is a continuous function with period \( 2\pi \), we can express it as a Fourier series:
   \[
   f(x) = \sum_{n=-\infty}^{\infty} d_n e^{inx},
   \]
   where \( i \) is the imaginary unit.

   Similarly, we know that:
   \[
   \sin(t) = \frac{e^{it} - e^{-it}}{2i}.
   \]

2. **Using the orthogonality of exponential functions:**

   For any integer \( m \),
   \[
   \int_0^{2\pi} e^{imx} \, dx = 2\pi \delta_m,
   \]
   where \( \delta_m \) is the Kronecker delta symbol, which is 1 if \( m = 0 \) and 0 otherwise.

3. **Evaluating the given integral:**

   We need to evaluate:
   \[
   \int_0^{2\pi} f(t - x) \sin(t) \, dt.
   \]
   Substituting the Fourier series for \( f(t - x) \):
   \[
   f(t - x) = \sum_{n=-\infty}^{\infty} d_n e^{in(t-x)} = \sum_{n=-\infty}^{\infty} d_n e^{in(t)} e^{-inx}.
   \]
   Therefore,
   \[
   \int_0^{2\pi} f(t - x) \sin(t) \, dt = \int_0^{2\pi} \left( \sum_{n=-\infty}^{\infty} d_n e^{in(t)} e^{-inx} \right) \sin(t) \, dt.
   \]

4. **Substituting \(\sin(t)\) and simplifying:**

   Substitute \(\sin(t) = \frac{e^{it} - e^{-it}}{2i}\):
   \[
   \int_0^{2\pi} \left( \sum_{n=-\infty}^{\infty} d_n e^{in(t)} e^{-inx} \right) \frac{e^{it} - e^{-it}}{2i} \, dt.
   \]
   Distribute the sum and integral:
   \[
   \sum_{n=-\infty}^{\infty} d_n e^{-inx} \int_0^{2\pi} e^{i(n+1)t} \frac{1}{2i} \, dt - \sum_{n=-\infty}^{\infty} d_n e^{-inx} \int_0^{2\pi} e^{i(n-1)t} \frac{1}{2i} \, dt.
   \]

5. **Using orthogonality to evaluate integrals:**

   Using the orthogonality property:
   \[
   \int_0^{2\pi} e^{i(n+1)t} \, dt = 2\pi \delta_{n+1,0} = 2\pi \delta_{n,-1},
   \]
   \[
   \int_0^{2\pi} e^{i(n-1)t} \, dt = 2\pi \delta_{n-1,0} = 2\pi \delta_{n,1}.
   \]
   Therefore, the integral simplifies to:
   \[
   \frac{d_{-1} e^{ix} - d_1 e^{-ix}}{2i}.
   \]

6. **Equating to \( cf(x) \):**

   Given that:
   \[
   \int_0^{2\pi} f(t - x) \sin(t) \, dt = cf(x),
   \]
   we have:
   \[
   \frac{d_{-1} e^{ix} - d_1 e^{-ix}}{2i} = c \sum_{n=-\infty}^{\infty} d_n e^{inx}.
   \]

7. **Solving for \( d_n \):**

   For the equality to hold, \( d_n \) must be zero for all \( |n| \neq 1 \). Thus, we have:
   \[
   d_{-1} = 2icd_1, \quad d_1 = -2icd_{-1}.
   \]
   Substituting \( d_{-1} \) into the second equation:
   \[
   d_1 = -2ic(2icd_1) = -4c^2 d_1.
   \]
   Therefore,
   \[
   1 = -4c^2 \implies c^2 = \frac{1}{4} \implies c = \frac{1}{2} \text{ or } c = -\frac{1}{2}.
   \]

8. **Finding \( f(x) \):**

   If \( c = \frac{1}{2} \), then:
   \[
   d_{-1} = i d_1.
   \]
   Thus,
   \[
   f(x) = d_1 (e^{ix} + i e^{-ix}) = d_1 (1 + i) (\cos(x) + \sin(x)).
   \]
   Given \( f(0) = 1 \):
   \[
   f(0) = d_1 (1 + i) = 1 \implies d_1 = \frac{1}{1+i} = \frac{1-i}{2}.
   \]
   Therefore,
   \[
   f(x) = \frac{1-i}{2} (1+i) (\cos(x) + \sin(x)) = \cos(x) + \sin(x).
   \]

The final answer is \( \boxed{ f(x) = \cos(x) + \sin(x) } \) and \( c = \frac{1}{2} \).